Question

a^3-1\a^3 -2a+2\a factorise

Answer 1

${a}^{3} - \frac{1}{ {a}^{3} } - 2a + \frac{2}{a}$

we already know that x^3 -y^3 =(x-y)(x^2-xy+y^2)

So we'll use this identity and then we're left with
$(a - \frac{1}{a} )( {a}^{2} + 1 + {1/a}^{2} ) - 2(a - \frac{1}{a} ) \\ =( a - \frac{1}{a})( {a}^{2} + 1 + {1/a}^{2} - 2)$
= {a-(1/a)} { a^2 -1 +(1/a^2)}

Answer 2

The answer is given below :

Now,

a³ - 1/a³ - 2a + 2/a

= (a - 1/a){a² + (a × 1/a) + 1/a²} - 2(a - 1/a)

= (a - 1/a)(a² + 1/a² + 1) - 2(a - 1/a)

= (a - 1/a)(a² + 1/a² + 1 - 2)

= (a - 1/a)(a² + 1/a² - 1),

which is the required factorization.

Thank you for your question.