Question 150

If $$x+y+z=6$$ and $$xy+yz+zx=10$$, then the value of $$x^3+y^3+z^3-3xyz$$ is:

Solution

Given : $$x+y+z=6$$ and $$xy+yz+zx=10$$ ---------------(i)

Now, $$(x+y+z)^2=(6)^2$$

=> $$x^2+y^2+z^2+2(xy+yz+zx)=36$$

=> $$x^2+y^2+z^2+2(10)=36$$

=> $$x^2+y^2+z^2=36-20=16$$ -----------------(ii)

To find : $$x^3+y^3+z^3-3xyz$$

= $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

= $$(6)(16-10)$$

= $$6\times6=36$$

=> Ans - (D)

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