Question

a cube minus one upon x minus 2 A + 2 upon 8​

Answer 1

$\boxed{\underline{\textsf{Identities :}}}$

$1.\:x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$

$2.\:x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$

$3.\:x^{2}-y^{2}=(x+y)(x-y)$

$4.\:x^{2}+y^{2}=(x+y)^{2}-2xy$

$5.\:(x+y)^{2}=x^{2}+2xy+y^{2}$

$6.\:(x-y)^{2}=x^{2}-2xy+y^{2}$

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:a^{3}-\frac{1}{a^{3}}-2a+\frac{2}{a}$

$=(a^{3}-\frac{1}{a^{3}})-(2a-\frac{2}{a})$

$=\small{(a-\frac{1}{a})\{a^{2}+(a*\frac{1}{a})+\frac{1}{a^{2}}\}-2(a-\frac{1}{a})}$

$=(a-\frac{1}{a})(a^{2}+1+\frac{1}{a^{2}})-2(a-\frac{1}{a})$

$=(a-\frac{1}{a})(a^{2}+1+\frac{1}{a^{2}}-2)$

$=(a-\frac{1}{a})(a^{2}+\frac{1}{a^{2}}-1)$

$\textsf{which is the required factorization.}$

Answer 2

Answer:

(a-1/a)(a²-1+1/a²)

Step-by-step explanation:

Hope it will helps you