Question 144

What will be the value of $$\ x^{3} \ + y^{3}+ \ z^{3}$$- 3xyz when x + y + z = 9 and $$\ x^{2} \ + y^{2} \ + z^{2}$$ =31 ?

Solution

Given : $$x+y+z=9$$ ---------(i)

and $$x^2+y^2+z^2=31$$ -----------(ii)

Squaring equation (i), we get :

=> $$(x+y+z)^2=(9)^2$$

=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=81$$

=> $$31+2(xy+yz+zx)=81$$

=> $$2(xy+yz+zx)=81-31=50$$

=> $$xy+yz+zx=\frac{50}{2}=25$$ -----------(iii)

To find : $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$

= $$(9)\times(31-25)$$

= $$9\times6=54$$

=> Ans - (C)

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