Question 108

If $$x + y + z = 1, xy + yz + zx = -26$$ and $$x^3 + y^3 + z^3 = 151$$, then what will be the value of $$xyz$$?

Solution

Given : $$xy + yz + zx = -26$$ and $$x^3 + y^3 + z^3 = 151$$ ------------(i)

Also, $$x + y + z = 1$$

Squaring both sides, => $$(x+y+z)^2=1$$

=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=1$$

=> $$(x^2+y^2+z^2)+2(-26)=1$$

=> $$(x^2+y^2+z^2)=1+52=53$$ --------------(ii)

We know that, $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

=> $$151-3xyz=(1)(53+26)$$

=> $$3xyz=151-79=72$$

=> $$xyz=\frac{72}{3}=24$$

=> Ans - (A)

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