If X + Y + Z = 9 then $$(X-4)^3+(Y-2)^3+(Z-3)^3-3(X- 4)(Y-2)(Z-3)$$ is
Given : $$X+Y+Z=9$$
=> $$(X-4)+(Y-2)+(Z-3)=0$$ ----------(i)
Let $$(X-4)=a , (Y-2)=b , (Z-3)=c$$
=> $$a+b+c=0$$ Â Â [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)
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