If 0 1 ( x 21 + x 14 + x 7 ) ( 2 x 14 + 3 x 7 + 6 ) 1 / 7 d x = 1 l ( 11 ) m where l , m , n , ? m and n are coprime, then l + m + n is equal to _______ . [2023]

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Solution

Q.
$If \int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m}$ $where l, m, n \in ℕ, m and n are coprime, then l + m + n is equal to$ _______ . [2023]

Ans.
(63)

$\int_{0}^{1} (x^{21} + x^{14} + x^{7}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x = \frac{1}{l} {(11)}^{m / n}, l, m, n \in ℕ$

L.H.S. $= \int_{0}^{1} x (x^{20} + x^{13} + x^{6}) {(2 x^{14} + 3 x^{7} + 6)}^{1 / 7} d x$

$= \int_{0}^{1} (x^{20} + x^{13} + x^{6}) {(2 x^{21} + 3 x^{14} + 6 x^{7})}^{1 / 7} d x$

Let $2 x^{21} + 3 x^{14} + 6 x^{7} = t^{7} \Rightarrow 42 (x^{20} + x^{13} + x^{6}) d x = 7 t^{6} d t$

$\Rightarrow (x^{20} + x^{13} + x^{6}) d x = \frac{t^{6}}{6} d t,$ When $x$ = 0, $t$ = 0 and when $x$ = 1,

$t = 11^{1 / 7}$ $\Rightarrow L.H.S. = \int_{0}^{11^{1 / 7}} (\frac{t^{6}}{6} \cdot t) d t$

$= \frac{1}{6} \int_{0}^{11^{1 / 7}} t^{7} d t$ $= \frac{1}{6} {[\frac{t^{8}}{8}]}_{0}^{{(11)}^{1 / 7}} = \frac{11^{8 / 7}}{48}$

Comparing with R.H.S., we get $l = 48, m = 8, n = 7$

$∴$ $l + m + n = 48 + 8 + 7 = 63$

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