Question 66

If X + Y + Z = 9 then $$(X-4)^3+(Y-2)^3+(Z-3)^3-3(X- 4)(Y-2)(Z-3)$$ is

Solution

Given : $$X+Y+Z=9$$

=> $$(X-4)+(Y-2)+(Z-3)=0$$ ----------(i)

Let $$(X-4)=a , (Y-2)=b , (Z-3)=c$$

=> $$a+b+c=0$$ [From (i)]

To find : $$(X-4)^3+(Y-2)^3+(Z-3)^3-3(X- 4)(Y-2)(Z-3)$$

= $$a^3+b^3+c^3-3abc$$

Using, $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

=> $$a^3+b^3+c^3-3abc=0 \times (a^2+b^2+c^2-ab-bc-ca)=0$$

=> Ans - (C)

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: