Question

${\left(a+\frac{1}{a}\right)}^2=3$ then $a^3+\frac{1}{a^3}=$

• A. $2\sqrt{2}$
• B. $\frac{4\sqrt{3}}{3}$
• C. $\frac{3\sqrt{3}}{4}$
• D. 0

Answer 1

Answer: (D)

0

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

Using identity,

$x^3+y^3=(x+y)(x^2-xy+y^2)a^3+\frac{1}{a^3}=(a+\frac{1}{a})(a^2-a\times \frac{1}{a}+\frac{1}{a^2})=(a+\frac{1}{a})(a^2+\frac{1}{a^2}-1)=(a+\frac{1}{a})(1-1)=(a+\frac{1}{a})\left(0\right)=0=>a^3+\frac{1}{a^3}=0Ans$