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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