Question

if p+q+r=9 then find (3-p)3 +(3-q)3 +(3-r)3
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Answer 1

$\textbf{Formula used:}$

$\text{If \bf\,a+b+c=0 , then \bf\,a^3+b^3+c^3=3abc }$

$\textbf{Given:}$

$p+q+r=9$

$\textbf{To find:}$

$(3-p)^3+(3-q)^3+(3-r)^3$

$\textbf{Solution:}$

$\text{Consider,}$

$p+q+r=9$

$\implies\,9-(p+q+r)=0$

$\implies\,(3-p)+(3-q)+(3-r)=0$

$\text{Using the above formula, we get}$

$(3-p)^3+(3-q)^3+(3-r)^3=3\,(3-p)(3-q)(3-r)$

$(3-p)^3+(3-q)^3+(3-r)^3=3[3^3-(p+q+r)3^2+(pq+qr+rp)3+pqr]$

$(3-p)^3+(3-q)^3+(3-r)^3=3[27-(9)3^2+3(pq+qr+rp)+pqr]$

$(3-p)^3+(3-q)^3+(3-r)^3=3[27-81+3(pq+qr+rp)+pqr]$

$\implies\bf(3-p)^3+(3-q)^3+(3-r)^3=3[pqr+3(pq+qr+rp)-54]$

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If cos theta +sec theta = root 3, prove that cos cube theta +sec cube theta = 0

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Answer 2

Answer:

3(3-p)(3-q)(3-r)

Step-by-step explanation:

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