Question

if x + y + z = 6 , xyz = 6 and xy + yz + zx = 11, then find x3+y3+z3​

Answer 1

According to the question,

It is given that,

$x + y + z = 6$

$xyz=6$

$xy + yz + zx = 11$

To find:- the value of $x^3+y^3+z^3$

By using the identity,

$(x+y+z)^3=x^3+y^3+z^3+3[(x+y+z)(xy+yz+xz)-(xyz)]...(1)$

Now, by substituting the given values in the equation (1) we get,

$=>(6)^3=x^3+y^3+z^3+3[6\times11-6]$

By simplifying the left side we get,

$=>216=x^3+y^3+z^3+3[6\times11-6]$

Multiply 6 with 11 we get,

$=>216=x^3+y^3+z^3+3(66-6)$

Now, subtract the bracket terms, we get

$=>216=x^3+y^3+z^3+3(60)$

Multiplying 3 with 60 we get,

$=>216=x^3+y^3+z^3+180$

$=>216-180=x^3+y^3+z^3$

$=>36=x^3+y^3+z^3$

Hence, the required value of $x^3+y^3+z^3$ is $36.$

Answer 2

Given,

$x+y+z=6$,

$xyz=6$,

And

$xy+yz+zx=11$

By using the $(x+y+z)^{3}$ identity to solve the given

$(x+y+z)^{3}=x^{3}+y^{3}+z^{3}+3[(x+y+z)(xy+yz+xz)-(xyz)]$

Substitute the given values in this identity

$(6)^{3}=x^{3}+y^{3}+z^{3}+3[(6)(11)-6]$

$216=x^{3}+y^{3}+z^{3}+3[66-6]$

$216=x^{3}+y^{3}+z^{3}+3\times60$

$216=x^{3}+y^{3}+z^{3}+180$

$x^{3}+y^{3}+z^{3}=216-180$

$x^{3}+y^{3}+z^{3}=36$

Therefore, the value of $x^{3}+y^{3}+z^{3}$ is $36$.