(3-p)^3 +(3-q)^3 +(3-r)^3=
Answers
\textbf{Formula used:}Formula used:
\text{If $\bf\,a+b+c=0$, then $\bf\,a^3+b^3+c^3=3abc$}If a+b+c=0, then a3+b3+c3=3abc
\textbf{Given:}Given:
p+q+r=9p+q+r=9
\textbf{To find:}To find:
(3-p)^3+(3-q)^3+(3-r)^3(3−p)3+(3−q)3+(3−r)3
\textbf{Solution:}Solution:
\text{Consider,}Consider,
p+q+r=9p+q+r=9
\implies\,9-(p+q+r)=0⟹9−(p+q+r)=0
\implies\,(3-p)+(3-q)+(3-r)=0⟹(3−p)+(3−q)+(3−r)=0
\text{Using the above formula, we get}Using the above formula, we get
(3-p)^3+(3-q)^3+(3-r)^3=3\,(3-p)(3-q)(3-r)(3−p)3+(3−q)3+(3−r)3=3(3−p)(3−q)(3−r)
(3-p)^3+(3-q)^3+(3-r)^3=3[3^3-(p+q+r)3^2+(pq+qr+rp)3+pqr](3−p)3+(3−q)3+(3−r)3=3[33−(p+q+r)32+(pq+qr+rp)3+pqr]
(3-p)^3+(3-q)^3+(3-r)^3=3[27-(9)3^2+3(pq+qr+rp)+pqr](3−p)3+(3−q)3+(3−r)3=3[27−(9)32+3(pq+qr+rp)+pqr]
(3-p)^3+(3-q)^3+(3-r)^3=3[27-81+3(pq+qr+rp)+pqr](3−p)3+(3−q)3+(3−r)3=3[27−81+3(pq+qr+rp)+pqr]
\implies\bf(3-p)^3+(3-q)^3+(3-r)^3=3[pqr+3(pq+qr+rp)-54]⟹(3−p)3+(3−q)3+(3−r)3=3[pqr+3(pq+qr+rp)−54]