Question : If $x+y+z = 9$, then the value of $(x−4)^3+(y−2)^3+(z−3)^3−3(x−4)(y−2)(z−3)$ is:
Option 1: 6
Option 2: 9
Option 3: 0
Option 4: 1
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Correct Answer: 0
Solution :
Given: $x+y+z=9$
Solution:
Let $(x–4) = a, (y−2) = b, (z−3) = c$
So, this equation stands to
$(x−4)^3+(y−2)^3+(z−3)^3−3(x−4)(y−2)(z−3)$
= $a^3+b^3+c^3–3abc$
Also, We know that $a^3+b^3+c^3–3abc$ = $(a+b+c)(a^2+b^2+c^2–bc–ab–ac) $.
So, $a+b+c = x−4+y−2+z−3$
⇒ $a+b+c = x+y+z−9$
⇒ $a+b+c = 9−9$
⇒ $a+b+c = 0$
So, $(x−4)^3+(y−2)^3+(z−3)^3−3(x−4)(y−2)(z−3)=0$
Hence, the correct answer is 0.
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