Question : If $x^3+y^3+z^3=3(1+xyz), P = y +z -x, Q = z+x-y$ and $R = x+y-z$, then what is the value of $P^3 + Q^3+R^3 - 3PQR?$
Option 1: 9
Option 2: 8
Option 3: 12
Option 4: 6
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Correct Answer: 12
Solution :
Given: $P=y+z-x,Q=z+x-y,R=x+y-z$ and $x^{3}+y^{3}+z^{3}-3xyz=3$
We know, $P^{3}+Q^{3}+R^{3}-3PQR=\frac{1}{2}(P+Q+R)((P-Q)^{2}+(Q-R)^{2}+(R-P)^{2})$
$P+Q+R=x+y+z$
$(P-Q)^{2}=(2y-2x)^2=4(x^2+y^2-2xy)$
$(Q-R)^{2}=(2z-2y)^2=4(z^2+y^2-2zy)$
$(R-P)^{2}=(2x-2z)^2=4(x^2+z^2-2xz)$
So, $P^{3}+Q^{3}+R^{3}-3PQR=\frac{1}{2}(x+y+z)(4(2x^2+2y^2+2z^2-2xy-2yz-2zx))$
⇒ $P^{3}+Q^{3}+R^{3}-3PQR=4(x^{3}+y^{3}+z^{3}-3xyz)$
$\therefore P^{3}+Q^{3}+R^{3}-3PQR=4\times 3=12$
Hence, the correct answer is 12.
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