৯ম – ১০ম শ্রেণির অনুশীলনী–৩.৩ : উৎপাদকে বিশ্লেষণ
\begin{align*}
&\mathbf{১. ab\left(x-y\right)-bc\left(x-y\right)}\\
&সমাধানঃ\\
&ab\left(x-y\right)-bc\left(x-y\right)\\
&=b\left\{a\left(x-y\right)-c\left(x-y\right)\right\} [কমন]\\
&=b\left(x-y\right)\left(a-c\right) [কমন]\\
&\mathbf{২. 9x^{2}+24x+16}\\
&সমাধানঃ\\
&9x^{2}+24x+16\\
&={\left(3x\right)}^{2}+2\cdot \left(3x\right)\cdot 4+4^{2}\\
&={\left(3x+4\right)}^{2} \ \ \ \ \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right]\\
&=(3x+4) (3x+4)\\
&\mathbf{৩. a^{4}-27a^{2}+1}\\
&সমাধানঃ\\
&a^{4}-27a^{2}+1\\
&=a^{4}-2a^{2}+1-25a^{2}\\
&={\left(a^{2}\right)}^{2}-2.a^{2}.1+1^{2}-25a^{2}\\
&={\left(a^{2}-1\right)}^{2}-{\left(5a\right)}^{2} [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&=\left(a^{2}-1+5a\right)\left(a^{2}-1-5a\right) \ \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right]\\
&=\left(a^{2}+5a-1\right)\left(a^{2}-5a-1\right)\\
&\mathbf{৪. x^{4}-6x^{2}y^{2}+y^{4}}\\
&সমাধানঃ\\
&x^{4}-6x^{2}y^{2}+y^{4}\\
&=x^{4}-2x^{2}y^{2}+y^{4}-4x^{2}y^{2}\\
&={\left(x^{2}-y^{2}\right)}^{2}-{\left(2xy\right)}^{2} [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&=\left(x^{2}-y^{2}+2xy\right)\ \left(x^{2}-y^{2}-2xy\right) \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right]\\
&= \left(x^{2}+2xy\ -y^{2}\right)\ \left(x^{2}-2xy\ -y^{2}\right) \\
&\mathbf{৫. \left(a^{2}-b^{2}\right)\left(x^{2}-y^{2}\right)+4abxy}\\
&সমাধানঃ\\
&\left(a^{2}-b^{2}\right)\left(x^{2}-y^{2}\right)+4abxy\\
&=a^{2}x^{2}-a^{2}y^{2}-b^{2}x^{2}+b^{2}y^{2}+4abxy\\
&=\left(a^{2}x^{2}+2abxy+b^{2}y^{2}\right)-\left(a^{2}y^{2}-2abxy+b^{2}x^{2}\right)\\
&={\left(ax+by\right)}^{2}-{\left(ay-bx\right)}^{2} \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right] [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&=\{\left(ax+by)+(ay-bx\right)\}\ \{\ \left(ax+by)-(ay\ -bx\right)\} \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right]\\
&= \left(ax+by+ay-bx\right)\ \left(ax+by-ay+bx\right)\\
&\mathbf{৬. 4a^{2}-12ab+9b^{2}-4c^{2}}\\
&সমাধানঃ\\
&4a^{2}-12ab+9b^{2}-4c^{2}\\
&= {\left(2a\right)}^{2}-2.2a.3b+{\left(3b\right)}^{2}-4c^{2}\\
&={\left(2a-3b\right)}^{2}-{\left(2c\right)}^{2} [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&=\left(2a-3b+2c\right)\left(2a-3b-2c\right) \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right]\\
\end{align*}
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\begin{align*}
&\mathbf{৭. a^{2}+6a+8-y^{2}+2y}\\
&সমাধানঃ\\
&a^{2}+6a+8-y^{2}+2y\\
&=a^{2}+6a+9-1-y^{2}+2y\\
&= a^{2}+2.a.3+3^{2}-\left(1+y^{2}-2y\right)\\
&={\left(a+3\right)}^{2}-\left(y^{2}-2.y.1+1^{2}\right) \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right] \\
&={\left(a+3\right)}^{2}-{\left(y-1\right)}^{2} [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&=\{\left(a+3)+(y-1\right)\}\ \{\left(a+3)-(y-1\right)\} \ \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&= \left(a+3+y-1\right)\ \left(a+3-y+1\right)\\
&=\left(a+y+2\right)\ \left(a-y+4\right)\\
&\mathbf{৮. 16x^{2}-25y^{2}-8xz+10yz}\\
&সমাধানঃ\\
&16x^{2}-25y^{2}-8xz+10yz\\
&= {\left(4x\right)}^{2}-{\left(5y\right)}^{2}-2z\left(4x-5y\right) [কমন]\\
&=\left(4x+5y\right)\left(4x-5y\right)-2z\left(4x-5y\right) \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&= \left(4x-5y\right)\left(4x+5y-2z\right)\\
&\mathbf{৯. x^{2}+13x+36}\\
&সমাধানঃ\\
&x^{2}+13x+36\\
&=x^{2}+9x+4x+36 [ মিডল টার্ম ব্রেক ] \\
&=x\left(x+9\right)+4\left(x+9\right) [কমন]\\
&=\left(x+4\right)\left(x+9\right) [কমন]\\
&\mathbf{১০. x^{4}+x^{2}-20}\\
&সমাধানঃ\\
&x^{4}+x^{2}-20\\
&=x^{4}+5x^{2}-4x^{2}-20 [ মিডল টার্ম ব্রেক ]\\
&=x^{2}\left(x^{2}+5\right)-4\left(x^{2}+5\right) [কমন]\\
&=\left(x^{2}+5\right)\left(x^{2}-4\right) [কমন]\\
&=\ \left(x^{2}-2^{2}\right)\left(x^{2}+5\right)\\
&=\left(x+2\right)\left(x-2\right)\left(x^{2}+5\right) \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&\mathbf{১১. a^{2}-30a+216}\\
&সমাধানঃ\\
&a^{2}-30a+216\\
&=a^{2}-18a-12a+216 [ মিডল টার্ম ব্রেক ]\\
&=a\left(a-18\right)-12\left(a-18\right) [কমন]\\
&=\left(a-18\right) a-12 [কমন]\\
&\mathbf{১২. a^{8}-a^{4}-2}\\
&সমাধানঃ\\
&a^{8}-a^{4}-2\\
&=a^{8}-2a^{4}+a^{4}-2 [ মিডল টার্ম ব্রেক ]\\
&=a^{4}\left(a^{4}-2\right)+1\left(a^{4}-2\right) [কমন]\\
&=\left(a^{4}-2\right)\left(a^{4}+1\right) [কমন]\\
&\mathbf{১৩. x^{2}-37x-650}\\
&সমাধানঃ\\
&x^{2}-37x-650\\
&=x^{2}-50x+13x-650 [ মিডল টার্ম ব্রেক ]\\
&=x\left(x-50\right)+13\left(x-50\right) [কমন]\\
&=\left(x-50\right)\ \left(x+13\right) [কমন]\\
&\mathbf{১৪. 9x^{2}y^{2}-5xy^{2}-14y^{2}}\\
&সমাধানঃ\\
&9x^{2}y^{2}-5xy^{2}-14y^{2}\\
&=y^{2}\left(9x^{2}-5x-14\right) [কমন]\\
&=y^{2}\left(9x^{2}-14x+9x-14\right) [ মিডল টার্ম ব্রেক ]\\
&=y^{2}\left\{x\left(9x-14\right)+1\left(9x-14\right)\right\} [কমন]\\
&=y^{2}\left(9x-14\right)\left(x+1\right) [কমন]\\
&\mathbf{১৫. 4x^{4}-27x^{2}-81}\\
&সমাধানঃ\\
&4x^{4}-27x^{2}-81\\
&=4x^{4}-36x^{2}+9x^{2}-81 [ মিডল টার্ম ব্রেক ]\\
&=4x^{2}\left(x^{2}-9\right)+9\left(x^{2}-9\right) [কমন]\\
&=\left(x^{2}-9\right)\left(4x^{2}+9\right) [কমন]\\
&= \left(x^{2}-3^{2}\right)\left(4x^{2}+9\right)\\
&=\left(x+3\right)\left(x-3\right)\left(4x^{2}+9\right) \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&\mathbf{১৬. ax^{2}+\left(a^{2}+1\right)x+a}\\
&সমাধানঃ\\
&ax^{2}+\left(a^{2}+1\right)x+a\\
&=ax^{2}+a^{2}x+x+a\\
&=ax\left(x+a\right)+1\left(x+a\right) [কমন]\\
&=\left(x+a\right)\left(ax+1\right) [কমন]\\
&\mathbf{১৭. 3{\left(a^{2}+2a\right)}^{2}-22\left(a^{2}+2a\right)+40}\\
&সমাধানঃ\\
&ধরি, a^{2}+2a=m\\
&3{\left(a^{2}+2a\right)}^{2}-22\left(a^{2}+2a\right)+40\\
&=3m^{2}-22m+40\\
&=3m^{2}-12m-10m+40 [ মিডল টার্ম ব্রেক ]\\
&=3m\left(m-4\right)-10\left(m-4\right) [কমন]\\
&=\left(m-4\right)\left(3m-10\right) [কমন]\\
&=\ \ \left(a^{2}+2a-4\right)\{3\left(a^{2}+2a\right)-10\} [ a^{2}+2a=m\ \ বসিয়ে\ ]\\
&=\left(a^{2}+2a-4\right)\left(3a^{2}+6a-10\right)\\
&\mathbf{১৮. \left(a-1\right)x^{2}+a^{2}xy+\left(a+1\right)y^{2}}\\
&সমাধানঃ\\
&\left(a-1\right)x^{2}+a^{2}xy+\left(a+1\right)y^{2}\\
&=\left(a-1\right)x^{2}+\left(a^{2}-1\right)xy+xy+\left(a+1\right)y^{2} \\
&[ মিডল টার্ম ব্রেকঃ \left(a+1\right)\left(a-1\right)=a^{2}-1 \text{ } তাহলে \text{ } পদ \text{ } দুইটি \text{ } a^{2}-1 এবং 1 \\
&অর্থাৎ \left(a^{2}-1\right)xy+xy\ =a^{2}xy-xy+xy=a^{2}xy]\\
&= \left(a-1\right)x^{2}+\left(a^{2}-1^{2}\right)xy+xy+\left(a+1\right)y^{2}\\
&=\left(a-1\right)x^{2}+\left(a+1\right)\left(a-1\right)xy+xy+\left(a+1\right)y^{2} \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&=\left(a-1\right)x\left\{x+\left(a+1\right)y\right\}+y\left\{x+\left(a+1\right)y\right\} [কমন]\\
&=\left\{x+\left(a+1\right)y\right\}\left\{\left(a-1\right)x+y\right\} [কমন]\\
&= \left(x+ay+y\right)\left(ax-x+y\right)\\
&\mathbf{১৯. x^{3}+3x^{2}+3x+2}\\
&সমাধানঃ\\
&x^{3}+3x^{2}+3x+2\\
&=x^{3}+3x^{2}+3x+1+1\\
&= x^{3}+3.x^{2}.1+3.x.1^{2}+1^{3}+1\\
&={\left(x+1\right)}^{3}+1^{3} [ সূত্র {\left(a+b\right)}^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} ]\\
&= \left(x+1+1\right)\{{\left(x+1\right)}^{2}-\left(x+1\right).1+1^{2}\} [ সূত্র a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right) ]\\
&=\left(x+2\right){\ \{\ x}^{2}+2.x.1+1^{2}-\left(x+1\right)+1\} \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right] \\
&= \left(x+2\right)\left(x^{2}+2x+1-x-1+1\right)\\
&=\left(x+2\right)\left(x^{2}+x+1\right)\\
&\mathbf{২০. a^{3}-6a^{2}+12a-9}\\
&সমাধানঃ\\
&a^{3}-6a^{2}+12a-9\\
&=a^{3}-6a^{2}+12a-8-1\\
&= {\ a}^{3}-3.a^{2}.2+3.a.2^{2}-2^{3}-1\\
&={\left(a-2\right)}^{3}-1^{3} [ সূত্র {\left(a+b\right)}^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} ]\\
&= \left(a-2-1\right){\ \{\ \left(a-2\right)}^{2}+\left(a-2\right).1+1^{2}\} [ সূত্র a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+\ ab+b^{2}\right) ]\\
&= \left(a-3\right)\left(a^{2}-2.a.2+2^{2}+a-2+1\right) [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}]\\
&= \left(a-3\right)\left(a^{2}-4a+4+a-1\right)\\
&=\left(a-3\right)\left(a^{2}-3a+3\right)\\
&\mathbf{২১. a^{3}-9b^{3}+{\left(a+b\right)}^{3}}\\
&সমাধানঃ\\
&a^{3}-9b^{3}+{\left(a+b\right)}^{3}\\
&= \ a^{3}-b^{3}+{\left(a+b\right)}^{3}-8b^{3}\\
&=a^{3}-b^{3}+{\left(a+b\right)}^{3}-{\left(2b\right)}^{3}\\
&=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)+\left(a+b-2b\right)\left\{{\left(a+b\right)}^{2}+2b\left(a+b\right)+4b^{2}\right\} [ সূত্র a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+\ ab+b^{2}\right) ]\\
&= \left(a-b\right)\left(a^{2}+ab+b^{2}\right)+\left(a-b\right)\left(a^{2}+2ab+b^{2}+2ab+2b^{2}+4b^{2}\right) \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right] \\
&= \left(a-b\right)\left(a^{2}+ab+b^{2}\right)+\left(a-b\right)\left(a^{2}+4ab+7b^{2}\right)\\
&=\left(a-b\right)\left\{a^{2}+ab+b^{2}+a^{2}+4ab+7b^{2}\right\} [কমন]\\
&=\left(a-b\right)\left(2a^{2}+5ab+8b^{2}\right)\\
&\mathbf{২২. 8x^{3}+12x^{2}+6x-63}\\
&সমাধানঃ\\
&8x^{3}+12x^{2}+6x-63\\
&= {\left(2x\right)}^{3}+3.{\left(2x\right)}^{2}.1+3.2x.1^{2}+1^{3}-64\\
&={\left(2x+1\right)}^{3}-64 [ সূত্র {\left(a+b\right)}^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} ]\\
&={\left(2x+1\right)}^{3}-4^{3}\\
&=\left(2x+1-4\right)\left\{{\left(2x+1\right)}^{2}+4\left(2x+1\right)+4^{2}\right\} [ সূত্র a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+\ ab+b^{2}\right) ]\\
&= \left(2x-3\right)\ \{{\left(2x\right)}^{2}+2.2x.1+1^{2}+8x+4+16)\} \left[সূত্র\ {\left(a+b\right)}^{2}=a^{2}+2ab+b^{2}\right] \\
&= \left(2x-3\right)\left(4x^{2}+4x+1+8x+20\right)\\
&=\left(2x-3\right)\left(4x^{2}+12x+21\right)\\
&\mathbf{২৩. 8a^{3}+\frac{b^{3}}{27}}\\
&সমাধানঃ\\
&8a^{3}+\frac{b^{3}}{27}\\
&={\left(2a\right)}^{3}+{\left(\frac{b}{3}\right)}^{3}\\
&= \left(2a+\frac{b}{3}\right){\{\left(2a\right)}^{2}-2a.\frac{b}{3}+{\left(\frac{b}{3}\right)}^{2}\} [ সূত্র a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right) ]\\
&=\left(2a+\frac{b}{3}\right)\left(4a^{2}-\frac{2ab}{3}+\frac{b^{2}}{9}\right) \\
&= \left(\frac{6a+b}{3}\right)\left(\frac{36a^{2}-6ab+b^{2}}{9}\right)\\
&=\frac{\left(6a+b\right)\left(36a^{2}-6ab+b^{2}\right)}{27}\\
&= \frac{1}{27}\left(6a+b\right)\left(36a^{2}-6ab+b^{2}\right) \\
&\mathbf{২৪. \frac{a^{6}}{27}-b^{6}}\\
&সমাধানঃ\\
&\frac{a^{6}}{27}-b^{6}\\
&={\left(\frac{a^{2}}{3}\right)}^{3}-{\left(b^{2}\right)}^{3}\\
&= \left(\frac{a^{2}}{3}-b^{2}\right){\{\left(\frac{a^{2}}{3}\right)}^{2}+\frac{a^{2}}{3}.b^{2}+{\left(b^{2}\right)}^{2}\} [ সূত্র a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+\ ab+b^{2}\right) ]\\
&=\left(\frac{a^{2}}{3}-b^{2}\right)\left(\frac{a^{4}}{9}+\frac{a^{2}b^{2}}{3}+b^{4}\right) \\
&= \left(\frac{a^{2}-3b^{2}}{3}\right)\left(\frac{a^{4}+3a^{2}b^{2}+9b^{4}}{9}\right)\\
&= \frac{\left(a^{2}-3b^{2}\right)\left(a^{4}+3a^{2}b^{2}+9b^{4}\right)}{27}\\
&=\frac{1}{27}\left(a^{2}-3b^{2}\right)\left(a^{4}+3a^{2}b^{2}+9b^{4}\right)\\
&\mathbf{২৫. 4a^{2}+\frac{1}{4a^{2}}-2+4a-\frac{1}{a}}\\
&সমাধানঃ\\
&4a^{2}+\frac{1}{4a^{2}}-2+4a-\frac{1}{a} \\
&=\ 4a^{2}-2+\frac{1}{4a^{2}}+4a-\frac{1}{a} \\
&= {\left(2a\right)}^{2}-2.2a.\frac{1}{2a}+\frac{1}{{\left(2a\right)}^{2}}+4a-\frac{1}{a} \\
&={\left(2a-\frac{1}{2a}\right)}^{2}+2\left(2a-\frac{1}{2a}\right) [সূত্র\ {\left(a-b\right)}^{2}=a^{2}-2ab+b^{2}] এবং [কমন]\\
&=\left(2a-\frac{1}{2a}\right)\left(2a-\frac{1}{2a}+2\right) [কমন]\\
&=\frac{4a^{2}-1}{2a}\cdot \frac{4a^{2}-1+4a}{2a}\ \\
&= \frac{{\left(2a\right)}^{2}-1^{2}}{2a}.\frac{4a^{2}+4a-1}{2a} \\
&=\frac{\left(2a+1\right)\left(2a-1\right)\left(4a^{2}+4a-1\right)}{4a^{2}} \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&\mathbf{২৬. {\left(3a+1\right)}^{3}-{\left(2a-3\right)}^{3}}\\
&সমাধানঃ\\
&{\left(3a+1\right)}^{3}-{\left(2a-3\right)}^{3}\\
&=\left\{\left(3a+1\right)-\left(2a-3\right)\right\}\left\{{\left(3a+1\right)}^{2}+\left(3a+1\right)\left(2a-3\right)+{\left(2a-3\right)}^{2}\right\}
[ সূত্র a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+\ ab+b^{2}\right) ]\\
&= \left(3a+1-2a+3\right)\{{\left(3a\right)}^{2}+2.3a.1+1^{2}+(3a.2a-3a.3+1.2a-1.3)+{\left(2a\right)}^{2}-2.2a.3+3^{2}\}\\
&=\left(a+4\right)\left\{\left(9a^{2}+6a+1\right)+\left(6a^{2}-9a\ +2a-3\right)+\left(4a^{2}-12a+9\right)\right\}\\
&=\left(a+4\right)\left(19a^{2}-13a+7\right)\\
&\mathbf{২৭. \left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-48}\\
&সমাধানঃ\\
&\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-48\\
&=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-48\\
&= \left(x^{2}+5x+2x+10\right)\left(x^{2}+4x+3x+12\right)-48\\
&=\left(x^{2}+7x+10\right)\left(x^{2}+7x+12\right)-48\\
&= \left(a+10\right)\left(a+12\right)-48 [ ধরি, x^{2}+7x=a]\\
&= a^{2}+12a+10a+120-48\\
&= a^{2}+22a+72\\
&= a^{2}+18a+4a+72 [ মিডল টার্ম ব্রেক]\\
&= a\left(a+18\right)+4\left(a+18\right)\\
&= \left(a+18\right)\left(a+4\right)\\
&=\left(x^{2}+7x+18\right)\left(x^{2}+7x+4\right) [ a=\ x^{2}+7x, বসিয়ে]\\
&\mathbf{২৮. \left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-65}\\
&সমাধানঃ\\
&\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-65\\
&=\left(x-1\right)\left(x-7\right)\left(x-3\right)\left(x-5\right)-65\\
&= \left(x^{2}-7x-x+7\right)\left(x^{2}-5x-3x+15\right)-65\\
&=\left(x^{2}-8x+7\right)\left(x^{2}-8x+15\right)-65\\
&= \left(a+7\right)\left(a+15\right)-65 [ ধরি, x^{2}-8x=a]\\
&= a^{2}+15a+7a+105-65\\
&= a^{2}+22a+40\\
&= a^{2}+20a+2a+40 [ মিডল টার্ম ব্রেক]\\
&= a\left(a+20\right)+2\left(a+20\right)\\
&= \left(a+20\right)\left(a+2\right)\\
&=\left(x^{2}-8x+20\right)\left(x^{2}-8x+2\right) [ a=\ x^{2}-8x\ , বসিয়ে]\\
&\mathbf{২৯.2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}}\\
&সমাধানঃ\\
&2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}\\
&= 4a^{2}b^{2}-a^{4}-b^{4}-c^{4}-2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}\\
&= 4a^{2}b^{2}-\left(a^{4}+b^{4}+c^{4}+2a^{2}b^{2}-2b^{2}c^{2}-2c^{2}a^{2}\right)\\
&= 4a^{2}b^{2}-\{{\left(a^{2}\right)}^{2}+{\left(b^{2}\right)}^{2}+{\left(-c^{2}\right)}^{2}+2a^{2}b^{2}+2b^{2}\left(-c^{2}\right)+2\left(-c^{2}\right)a^{2}\}\\
&= 4a^{2}b^{2}-{\{a^{2}+b^{2}+\left(-c^{2}\right)\}}^{2} [ সূত্র, {\left(a+b+c\right)}^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca]\\
&=4a^{2}b^{2}-{\left(a^{2}+b^{2}-c^{2}\right)}^{2} \\
&={\left(2ab\right)}^{2}-{\left(a^{2}+b^{2}-c^{2}\right)}^{2}\\
&=\left\{2ab+\left(a^{2}+b^{2}-c^{2}\right)\right\}\left\{2ab-\left(a^{2}+b^{2}-c^{2}\right)\right\} \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&= \{\left(a^{2}+2ab+b^{2}\right)-c^{2}\}\left(2ab-a^{2}-b^{2}+c^{2}\right)\\
&= \{{\left(a+b\right)}^{2}-c^{2}\}\left(c^{2}-a^{2}+2ab-b^{2}\right)\\
&= {\{\left(a+b\right)}^{2}-c^{2}\}\{c^{2}-\left(a^{2}-2ab+b^{2}\right)\}\\
&=\left\{{\left(a+b\right)}^{2}-c^{2}\right\}\left\{c^{2}-{\left(a-b\right)}^{2}\right\}\\
&=\left\{a+b+c\right\}\left\{a+b-c\right\}\left\{c+\left(a-b\right)\right\}\left\{c-\left(a-b\right)\right\} \left[সূত্র\ {\ \ a}^{2}-b^{2}=\left(a+b\right)\left(a-b\right)\right] \\
&=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\\
&\mathbf{৩০. 14{\left(x+z\right)}^{2}-29\left(x+z\right)\left(x+1\right)-15{\left(x+1\right)}^{2}}\\
&সমাধানঃ\\
&14{\left(x+z\right)}^{2}-29\left(x+z\right)\left(x+1\right)-15{\left(x+1\right)}^{2}\\
&= 14a^{2}-29ab-15b^{2} [ ধরি, x+z=a, এবং x+1=b]\\
&= 14a^{2}-35ab+6ab-15b^{2} [ মিডল টার্ম ব্রেক]\\
&= 7a\left(2a-5b\right)+3b\left(2a-5b\right)\\
&= \left(2a-5b\right)\left(7a+3b\right)\\
&=\left\{2\left(x+z\right)-5\left(x+1\right)\right\}\left\{7\left(x+z\right)+3\left(x+1\right)\right\} [ a=x+z, এবং b=x+1 বসিয়ে ]\\
&= \left(2x+2z-5x-5\right)\left(7x+7z+3x+3\right)\\
&=\left(2z-3x-5\right)\left(10x+7z+3\right)\\
&\mathbf{৩১. দেখাও যে, \left(x+1\right)\left(x+2\right)\left(3x-1\right)\left(3x-4\right)=\left(3x^{2}+2x-1\right)\left(3x^{2}+2x-8\right)}\\
&সমাধানঃ\\
&এখানে,\ \ \ \left(x+1\right)\left(3x-1\right)\\
&=3x^{2}-x+3x-1\\
&=3x^{2}+2x-1\\
&এবং\ \ \left(x+2\right)\left(3x-4\right)\\
&=3x^{2}-4x+6x-8\\
&=3x^{2}+2x-8\\
&সুতরাং \\
&\left(x+1\right)\left(x+2\right)\left(3x-1\right)\left(3x-4\right)\\
&=\left(x+1\right)\left(3x-1\right)\left(x+2\right)\left(3x-4\right)\\
&=\left(3x^{2}+2x-1\right)\left(3x^{2}+2x-8\right)\\
&তাহলে, \left(x+1\right)\left(x+2\right)\left(3x-1\right)\left(3x-4\right)=\left(3x^{2}+2x-1\right)\left(3x^{2}+2x-8\right)\\
&(দেখানো \text{ } হল)
\end{align*}