If $p + q = 2, p q = 1$ then the value of $p^{3} + q^{3} + 6 p^{2} + 5 q^{2} + 6 p q$ is
(Given $p$ and $q$ are rational numbers)

19

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18

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15

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23

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Solution

The correct option is C $19$
Given, $p + q = 2$ and $p q = 1$ ,
So, ${(p - q)}^{2} = {(p + q)}^{2} - 4 p q \Rightarrow {(p - q)}^{2} = 2^{2} - 4 = 0 \Rightarrow (p - q) = 0$
or $p = q$ As, $p + q = 2 \Rightarrow p = q = 1$
Hence, $p^{3} + q^{3} + 6 p^{2} + 5 q^{2} + 6 p q = 1 + 1 + 6 + 5 + 6 = 19$

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