Question

if x+2y=5, then find the value of x^3+8y^3+30xy-125​

Answer 1

Answer:

0 is the correct answer.

☸Given☸

• x+y = 5   --------------------------(1)

➺To Find:

• $\sf{x^{3}+8y^{3}+30xy-125}$

✯Solution✯

We know that,

• $\bf{(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)}$

So,

On cubing equation (1), we get

$\longrightarrow\sf{(x+y)^{3}={5}^{3}}$

$\longrightarrow\sf{x^{3}+(2y)^{3}+3(x)(2y)(x+2y)={125}}$

$\longrightarrow\sf{x^{3}+8y^{3}+6xy{\red{(x+2y)}}={125}}$

Now after feeding values of equation (1), we get

$\longrightarrow\sf{x^{3}+8y^{3}+6xy{\red{(5)}}={125}}$

$\longrightarrow\sf{x^{3}+8y^{3}+30xy={125}}$

$\longrightarrow\sf{x^{3}+8y^{3}+30xy-{125}=0}$

Hence, value of $\tt{\purple {{x}^{3}+{8y}^3+30xy-125}}$ is $\tt{\green{0}}$.