Question

(x+y)^1/3 +(y+z)^1/3 + (z+x)^1/3 =0 then find x^3+y^3+z^3

Answer 1

Answer:

$\frac{3}{8}(x+y)(y+z)(z+x)$

Step-by-step explanation:

We are given, $(x+y)^{\frac{1}{3}} +(y+z)^{\frac{1}{3}}+(z+x)^{\frac{1}{3}}=0$ ..... (1)

And, we have to find the value of (x³+y³+z³)

Now, from (1), we can write,

$(x+y)^{\frac{1}{3}} +(y+z)^{\frac{1}{3}}=-(z+x)^{\frac{1}{3}}$ .... (2)

Now, cubing both sides we get,

⇒$(x+y)+(y+z)+3(x+y)^{\frac{1}{3}}(y+z)^{\frac{1}{3}}[(x+y)^{\frac{1}{3}}+(y+z)^{\frac{1}{3}}]=-(z+x)$

⇒ $2(x+y+z)+3(x+y)^{\frac{1}{3}}(y+z)^{\frac{1}{3}}[-(z+x)^{\frac{1}{3}}]=0$ {from equation (2)}

⇒$2(x+y+z)= 3(x+y)^{\frac{1}{3}}(y+z)^{\frac{1}{3}}(z+x)^{\frac{1}{3}}$

Again cubing both sides we get,

⇒ $8(x+y+z)^{3}=27(x+y)(y+z)(z+x)$

We have the formula, (a+b+c)³=(a³+b³+c³)+3(a+b)(b+c)(c+a)

⇒ $8[(x^{3}+y^{3}+z^{3})+3(x+y)(y+z)(z+x)]=27(x+y)(y+z)(z+x)$

⇒ $8(x^{3}+y^{3}+z^{3})=3(x+y)(y+z)(z+x)$

⇒ $(x^{3}+y^{3}+z^{3})=\frac{3}{8}(x+y)(y+z)(z+x)$ (Answer)