$8 x^{3}-y^{3}+1-6 x y$
$=(2 x)^{3}+(-y)^{3}+(1)^{3}-3 \cdot 2 x \cdot(-y) \cdot 1$
$=(2 x-y+1)\left\{(2 x)^{2}+(-y)^{2}+(1)^{2}\right. -2 x \cdot(-y)-(-y) \cdot 1-1.2 x\}$
$=(2 x-y+1)\left(4 x^{2}+y^{2}+1+2 x y+y-2 x\right)$

$8 a^{3}-27 b^{3}-1-18 a b$
$=(2 a)^{3}+(-3 b)^{3}+(-1)^{3}-3 \cdot 2 a \cdot(-3 b)(-1)$
$=(2 a-3 b-1)\left\{(2 a)^{2}+(-3 b)^{2}+(-1)^{2}\right. -2 a \cdot(-3 b)-(-3 b) \cdot(-1)-(-1) \cdot 2 a\}$
$=(2 a-3 b-1)\left(4 a^{2}+9 b^{2}+1+6 a b-3 b+2 a\right)$

$1+8 x^{3}+18 x y-27 y^{3}$
$=(1)^{3}+(2 x)^{3}+(-3 y)^{3}-3 \cdot 1 \cdot 2 x \cdot(-3 y)$
$=(1+2 x-3 y)\left\{(1)^{2}+(2 x)^{2}+(-3 y)^{2}\right. -1.2 x-2 x \cdot(-3 y)-(-3 y) \cdot 1\}$
$=(1+2 x-3 y)\left(1+4 x^{2}+9 y^{2}-2 x+6 x y+3 y\right)$

$x^{3}+y^{3}-12 x y+64$
$=(x)^{3}+(y)^{3}+(4)^{3}-3 \cdot x \cdot y \cdot 4$
$=(x+y+4)\left\{(x)^{2}+(y)^{2}+(4)^{2}-x \cdot y-y \cdot 4-4 \cdot x\right\}$
$=(x+y+4)\left(x^{2}+y^{2}+16-x y-4 y-4 x\right)$

$(3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}$
$=(3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}-3(3 a-2 b)(2 b-5 c)(5 c-3 a)$
$\quad+3(3 a-2 b)(2 b-5 c)(5 c-3 a)$
$=(3 a-2 b+2 b-5 c+5 c-3 a)\left\{(3 a-2 b)^{2}+\right. (2 b-5 c)^{2}+(5 c-3 a)^{2}-(3 a-2 b)(2 b-5 c)$
$\quad-(2 b-5 c)(5 c-3 a)-(5 c-3 a)(3 a-2 b)\} +3(3 a-2 b)(2 b-5 c)(5 c-3 a)$
$=0 \times\{(3 a-2 b)^{2}+(2 b-5 c)^{2}+(5 c-3 a)^{2}-(3 a-2 b)(2 b-5 c)$
$\quad-(2 b-5 c)(5 c-3 a)-(5 c-3 a) (3 a-2 b)\}+3(3 a-2 b)(2 b-5 c)(5 c-3 a)$
$=3(3 a-2 b)(2 b-5 c)(5 c-3 a)$
বিকল্প পদ্ধতি :
ধরি, $3 a-2 b=x, 2 b-5 c=y$ এবং $5 c-3 a=z$
$\therefore x+y+z=3 a-2 b+2 b-5 c+5 c-3 a$
বা, $x+y+z=0$
বা, $x+y=-z$
বা, $(x+y)^{3}=(-z)^{3}$
বা, $x^{3}+y^{3}+3 x y(x+y)=-z^{3}$
বা, $x^{3}+y^{3}+z^{3}=-3 x y(-z)\quad[\because x+y=-z]$
বা, $x^{3}+y^{3}+z^{3}=3 x y z$
$\therefore (3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}=3(3 a-2 b)(2 b-5 c)(5 c-3 a)$

$(2 x-y)^{3}-(x+y)^{3}+(2 y-x)^{3}$
$=(2 x-y)^{3}+(-x-y)^{3}+(2 y-x)^{3}$
ধরি, $2 x-y=a,-x-y=b$ এবং $2y-x=c$
$\therefore a+b+c=2 x-y-x-y+2 y-x$
বা, $a+b+c=0$
বা, $a+b=-c$
বা, $(a+b)^{3}=(-c)^{3}$
বা, $a^{3}+b^{3}+3 a b(a+b)=-c^{3}$
বা, $a^{3}+b^{3}+c^{3}=-3 a b(-c) \quad[\because a+b=-c]$
বা, $a^{3}+b^{3}+c^{3}=3 a b c$
বা, $(2 x-y)^{3}+(-x-y)^{3}+(2 y-x)^{3} =3(2 x-y)(-x-y)(2 y-x)$
$\therefore(2 x-y)^{3}-(x+y)^{3}+(2 y-x)^{3} =-3(2 x-y)(x+y)(2 y-x)$

$a^{6}+32 a^{3}-64$
$=\left(a^{2}\right)^{3}+(2 a)^{3}+(-4)^{3}-3 \cdot a^{2} \cdot 2 a(-4)$
$=\left(a^{2}+2 a-4\right)\left\{\left(a^{2}\right)^{2}+(2 a)^{2}+(-4)^{2}-a^{2} \cdot 2 a-2 a \cdot(-4)-(-4) \cdot a^{2}\right\}$
$=\left(a^{2}+2 a-4\right)\left(a^{4}+4 a^{2}+16-2 a^{3}+8 a+4 a^{2}\right)$
$=\left(a^{2}+2 a-4\right)\left(a^{4}-2 a^{3}+8 a^{2}+8 a+16\right)$

$a^{6}-18 a^{3}+125$
$=\left(a^{2}\right)^{3}+(3 a)^{3}+(5)^{3}-3 \cdot a^{2} \cdot 3 a \cdot 5$
$=\left(a^{2}+3 a+5\right)\left\{\left(a^{2}\right)^{2}+(3 a)^{2}+(5)^{2}-a^{2} \cdot 3 a-3 a \cdot 5-5 \cdot a^{2}\right\}$
$=\left(a^{2}+3 a+5\right)\left(a^{4}+9 a^{2}+25-3 a^{3}-15 a-5 a^{2}\right)$
$=\left(a^{2}+3 a+5\right)\left(a^{4}-3 a^{3}+4 a^{2}-15 a+25\right)$

$p^{3}(q-r)^{3}+q^{3}(r-p)^{3}+r^{3}(p-q)^{3}$
$(p q-p r)^{3}+(q r-p q)^{3}+(p r-q r)^{3}$
ধরি, $p q-p r=x, q r-p q=y$ এবং $pr-qr=z$
$\therefore x+y+z=p q-p r+q r-p q+p r-q r$
বা, $x+y+z=0$
বা, $x+y=-z$
বা, $(x+y)^{3}=(-z)^{3}$
বা, $x^{3}+y^{3}+3 x y(x+y)=-z^{3}$
বা, $x^{3}+y^{3}+z^{3}=-3 x y(-z) \quad[\because x+y=-z]$
বা, $x^{3}+y^{3}+z^{3}=3 x y z$
বা, $(p q-p r)^{3}+(q r-p q)^{3}+(p r-q r)^{3}=3(p q-p r)(q r-p q)(p r-q r)$
$\therefore p^{3}(q-r)^{3}+q^{3}(r-p)^{3}+r^{3}(p-q)^{3} =3 p q r(q-r)(r-p)(p-q)$

$p^{3}+\frac{1}{p^{3}}+\frac{26}{27}$
$=p^{3}+\frac{1}{p^{3}}+\frac{27-1}{27}$
$=p^{3}+\frac{1}{p^{3}}+\frac{27}{27}-\frac{1}{27}$
$=p^{3}+\frac{1}{p^{3}}+1-\frac{1}{27}$
$=(p)^{3}+\left(\frac{1}{p}\right)^{3}+\left(-\frac{1}{3}\right)^{3}-3 \cdot p \cdot \frac{1}{p} \cdot\left(-\frac{1}{3}\right)$
$=\left(p+\frac{1}{p}-\frac{1}{3}\right)\left\{(p)^{2}+\left(\frac{1}{p}\right)^{2}+\left(-\frac{1}{3}\right)^{2}-p \cdot \frac{1}{p}-\frac{1}{p} \cdot\left(-\frac{1}{3}\right)-\left(-\frac{1}{3}\right) \cdot p\right\}$
$=\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^{2}+\frac{1}{p^{2}}+\frac{1}{9}-1+\frac{1}{3 p}+\frac{p}{3}\right)$
$=\left(p+\frac{1}{p}-\frac{1}{3}\right)\left(p^{2}+\frac{1}{p^{2}}-\frac{8}{9}+\frac{1}{3 p}+\frac{p}{3}\right)$