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Thenosic · Answer 1 · 2023-12-03T18:38:05+0000

$\begin{gathered} (4x-3)^4 \\ \end{gathered}$

This can be solved by binomial expansion and also by Pascal triangle coefficients.

Alternatively, it could be solved by the expansion of brackets.

$\begin{gathered} (4x-3)^4=(4x-3)(4x-3)(4x-3)(4x-3) \\ \lbrack(4x-3)(4x-3)\rbrack^2 \\ \end{gathered}$
$\begin{gathered} (4x-3)(4x-3)=16x^2-12x-12x+9 \\ =16x^2-24x+9 \end{gathered}$
$\begin{gathered} \lbrack(4x-3)(4x-3)\rbrack^2=(16x^2-24x+9)^2 \\ (16x^2-24x+9)(16x^2-24x+9)=256x^4-384x^3+144x^2-384x^3+576x^2-216x+144x^2-216x+81 \\ C\text{ ollecting the like terms,} \\ =256x^4-384x^3-384x^3+144x^2+576x^2+144x^2-216x-216x+81 \\ =256x^4-768x^3+864x^2-432x+81 \end{gathered}$

Therefore, the expansion is;

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