Question

Express a^3-1/a^3 - 3(a-1/a) as a perfect cube and find its value when a = 3

Answer 1

GIVEN :

Express as a perfect cube and find its value when a = 3

TO FIND :

Express as a perfect cube and find its value when a = 3

SOLUTION :

Given expression is $a^3-\frac{1}{a^3} - 3(a-\frac{1}{a})$

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

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

By using the algebraic identity :

$(a-b)^3=(a-b)(a^2+ab+b^2)$

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

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

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

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

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

By using the algebraic identity :

$(a-b)^2=a^2+b^2-2ab$

By using the exponent property :

$a^m.a^n=a^{m+n}$

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

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

⇒ the given expression $a^3-\frac{1}{a^3} - 3(a-\frac{1}{a})$ is expressed as a perfect cube is  $(a-\frac{1}{a})^3$

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

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

$=(\frac{9-1}{3})^3$

$=(\frac{8}{3})^3$

By using the exponent property :

$(\frac{a}{b})^m=\frac{a^m}{b^m}$

$=\frac{8^3}{3^3}$

$=\frac{512}{27}$

⇒ The value of $(a-\frac{1}{a})^3$ when a=3 is $\frac{512}{27}$