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

1.	Express a^3-1/a^3 - 3(a-1/a) as a perfect cube and find its value when a = 3
Answer» 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 $\frac{512}{27}$