Expand ( x ^ 2 + 3 x ) ^ 4 , x 0 - Vidyadip

Question

Expand $\left( x ^ { 2 } + \frac { 3 } { x } \right) ^ { 4 }$, x $\neq$ 0

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Answer

By using binomial theorem, we have
$(a+b)^n={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots+{ }^n C_r a^{n-r} b^r+\ldots+{ }^n C_n b^n$
$\therefore\left(x^2+\frac{3}{x}\right)^4={ }^4 C_0\left(x^2\right)^4+{ }^4 C_1\left(x^2\right)^{4-1}\left(\frac{3}{x}\right)^1+{ }^4 C_2\left(x^2\right)^{4-2}\left(\frac{3}{x}\right)^2+{ }^4 C_3\left(x^2\right)^{4-3}\left(\frac{3}{x}\right)^3+4 C_4\left(x^2\right)^{4-4}\left(\frac{3}{x}\right)^4$
$=1 \cdot x^8+4\left(x^2\right)^3\left(\frac{3}{x}\right)+6\left(x^2\right)^2\left(\frac{3}{x}\right)^2+4\left(x^2\right)\left(\frac{3}{x}\right)^3+1 \cdot\left(x^2\right)^0\left(\frac{3}{x}\right)^4$
${\left[\because{ }^4 C_0=1,{ }^4 C_1=4,{ }^4 C_2=6,{ }^4 C_3=4 \text { and }{ }^4 C_4=1\right]}$
$=x^8+4 \cdot x^6 \cdot \frac{3}{x}+6 x^4 \cdot \frac{9}{x^2}+4 x^2 \cdot \frac{27}{x^3}+1 \cdot 1 \cdot \frac{81}{x^4}$
$=x^8+12 x^5+54 x^2+\frac{108}{x}+\frac{81}{x^4}$

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