If $x+y+z=7$, $x^2+y^2+z^2=85$ and $x^3+y^3+z^3=913$, then value of $\sqrt[3]{xyz}$ is
Solution
✅ Correct Option: 3
Using the identity $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$: $7^2=85+2(xy+yz+zx)$ $\Rightarrow 49-85=2(xy+yz+zx)$ $\Rightarrow xy+yz+zx=-18$ Now using the identity $x^3+y^3+z^3-3xyz=(x+y+z)\left(x^2+y^2+z^2-xy-yz-zx\right)$: $913-3xyz=7\left(85-(-18)\right)$ $\Rightarrow 913-3xyz=7\times103=721$ $\Rightarrow 3xyz=192$ $\Rightarrow xyz=64$ $\sqrt[3]{xyz}=\sqrt[3]{64}=4$
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