If $$a + \frac{1}{a} = 3,  then   \left(a^4 + \frac{1}{a^4}\right)$$ is equal to:

Given

$$a + \frac{1}{a}$$ = 3

using the formula $$(a+b)^2$$ = $$a^2 + b^2+ 2 × a × b $$

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

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

$$\left(a^2 + \frac{1}{a^2}\right)$$ = 9 - 2

$$\left(a^2 + \frac{1}{a^2}\right)$$ = 7

again use the same formula to the above solutions we can get final answer $$(a+b)^2$$ = $$a^2 + b^2+ 2 × a × b $$

 $$\left(a^2 + \frac{1}{a^2}\right)^2$$ = $$\left(a^4 + \frac{1}{a^4}\right)$$+ $$ 2 $$ × $$a^2 × \frac{1}{a^2}$$

$$7^2$$ =  $$\left(a^4 + \frac{1}{a^4}\right)$$ + 2

$$49-2$$ = $$\left(a^4 + \frac{1}{a^4}\right)$$

therefore $$\left(a^4 + \frac{1}{a^4}\right)$$ = $$47$$