Question

(ix) If p=x+y=z, q=y+z=x, g=z+x-y then prove that
p3+q3+p3-4(x3+y3+23) = 3(pqr-4xyz)​

Answer 1

Answer:

p^3+q^3+r^3–3pqr

=( p+q+r)(1/2((p-q)^2+(q-r)^2+(r-p)^2))

=(x+y+z)(1/2*4*((x-y)^2 +(y-z^2+(z-x)^2)

= 4(x^3+y^3+z^3–3xyz)

=4(3(1+xyz)-3xyz)

=12

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