If $(3 x - 1)^{4} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$ , then find the value of $a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}$

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Solution

$(3 x - 1)^{4} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
$=> (3 x - 1)^{2} (3 x - 1)^{2} = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
$=> (9 x^{2} - 6 x + 1) (9 x^{2} - 6 x + 1) = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
$=> (81 x^{4} - 54 x^{3} + 9 x^{2} - 54 x^{3} - 6 x + 9 x^{2} - 6 x + 1) = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
$=> (81 x^{4} - 108 x^{3} + 18 x^{2} - 12 x + 1) = a_{4} x^{4} + a_{3} x^{3} + a_{2} x^{2} + a_{1} x + a_{0}$
Thus, $a_{4} = 81$
$a_{3} = - 108$
$a_{2} = 18$
$a_{1} = - 12$
$a_{0} = 1$
Now,
$a_{4} + 3 a_{3} + 9 a_{2} + 27 a_{1} + 81 a_{0}$
$= 81 + 3 (- 108) + 9 (18) + 27 (- 12) + 81 (1)$
$= 81 - 324 + 162 - 324 + 81$
$= 0$

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