Question

If (p²+1) (q²+1) +36 =12 (p+q) then (p^3 +q^3 ) =​

Answer 1

SOLUTION

GIVEN

$\sf{( {p}^{2} + 1)( {q}^{2} + 1) + 36 = 12(p + q)}$

TO DETERMINE

The value of

$\sf{ {p}^{3} + {q}^{3} }$

EVALUATION

$\sf{( {p}^{2} + 1)( {q}^{2} + 1) + 36 = 12(p + q)}$

$\sf{ \implies \: {p}^{2} {q}^{2} + {p}^{2} + {q}^{2} + 1+ 36 = 12(p + q)}$

$\sf{ \implies \: {p}^{2} + {q}^{2} +12(p + q) - 2pq + 36 + {p}^{2} {q}^{2} - 2pq + 1= 0}$

$\sf{ \implies \: {(p + q - 6)}^{2} + {(pq - 1)}^{2} = 0}$

We know if the sum of the squares of two real numbers are zero then they are separately zero

$\sf{ {(p + q - 6)}^{2} = 0 \: \: and \: \: {(pq - 1)}^{2} = 0}$

$\sf{ \implies \: {(p + q - 6)} = 0 \: \: and \: \: {(pq - 1)} = 0}$

$\sf{ \implies \: p + q = 6 \: \: and \: \:pq = 1}$

$\sf{ {p}^{3} + {q}^{3} }$

$\sf{ = {(p + q)}^{3} - 3pq(p + q) }$

$\sf{ = {(6)}^{3} - 3 \times 6 \times 1 }$

$\sf{ = 216 - 18 }$

$\sf{ = 198 }$

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