Question

y+1/4y=2 then 16y³+1/4y³=
​

Answer 1

Answer : 32 ✔✔

___________________________

Solution :

Given that :

$y + \frac{1}{4y} = 2$

As we know that :

${a}^{3} + {b}^{3} = {(a + b)}^{3} - 3ab(a + b)$

$= > {(y)}^{3} + \frac{1}{ {(4y)}^{3} } = {(y + \frac{1}{4y} )}^{3} - 3y. \frac{1}{y} (y + \frac{1}{4y} ) \\ \\ = > {y}^{3} + \frac{1}{64 {y}^{3} } = {2}^{3} - 3 \times 1(2) \\ \\ = > {y}^{3} + \frac{1}{64 {y}^{3} } = 8 - 6 = 2$

On multiplying by 16 in both the sides :

$16( {y}^{3} + \frac{1}{64 {y}^{3} }) = 16 \times 2 \\ \\ = > 16 {y}^{3} + \frac{1}{4 {y}^{3} } = 32$

____________________________

I hope it will help you ☺

Fóllòw Më ❤