Expand ( x 2 + 3 x ) 4 , x ? 0 .

Question

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

Smart Achievers · Accepted Answer

(1)

$We have,$

${(x^{2} + \frac{3}{x})}^{4} = {}^{4}C_{0} {(x^{2})}^{4} + {}^{4}C_{1} {(x^{2})}^{3} (\frac{3}{x}) + {}^{4}C_{2} {(x^{2})}^{2} {(\frac{3}{x})}^{2}$ $+ {}^{4}C_{3} (x^{2}) {(\frac{3}{x})}^{3} + {}^{4}C_{4} {(\frac{3}{x})}^{4}$

$= x^{8} + 4 x^{6} \cdot \frac{3}{x} + 6 x^{4} \cdot \frac{9}{x^{2}} + 4 x^{2} \cdot \frac{27}{x^{3}} + \frac{81}{x^{4}}$

$= x^{8} + 12 x^{5} + 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$

Solution

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

1 $x^{8} + 12 x^{5} + 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$

2 $x^{8} - 12 x^{5} + 54 x^{2} - \frac{108}{x} - \frac{81}{x^{4}}$

3 $x^{8} - 12 x^{5} - 54 x^{2} + \frac{108}{x} + \frac{81}{x^{4}}$

4 $None of these$

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Solution

Q. Expand (x2+3x)4, x≠0.

1 x8+12x5+54x2+108x+81x4

2 x8-12x5+54x2-108x-81x4

3 x8-12x5-54x2+108x+81x4

4 None of these