Question

If a is positive and $a^2+\frac{1}{a^2}=7$, thenfind $a^3+\frac{1}{a^3}$.

• A. 21
• B. $$3\sqrt{7}$$
• C. 18
• D. $$7\sqrt{7}$$

Answer 1

Answer:

C. 18

Solution:

Formula used:

(a + b)² = a² + 2ab + b²

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

Given that, $a^2+\frac{1}{a^2}=7$

$⇒a^2+\frac{1}{a^2}+2=9$

By using the above formula,

$⇒(a^2+\frac{1}{a^2})^2=9$

$⇒(a+\frac{1}{a})=3$ ----------(1)

We know  that, (a + b)³ = a³ + b³ + 3ab(a + b)

$⇒a^3+\frac{1}{a^3}+3(a\times{\frac{1}{a}})(a+\frac{1}{a}) =27$

Using equation (1)

$⇒a^3+\frac{1}{a^3}+3\times{3}=27$

$⇒a^3+\frac{1}{a^3}=27-9=18$