Question

Expand $\rm \Big(\frac{2x}{3} - \frac{3}{2x} \Big)^{4}$.

Answer 1

Answer:

16x⁴/81 - 16x²/9 + 6 - 9/x² + 81/(16x⁴)

Step-by-step explanation:

Hi,

Consider the expansion  (2x/3 - 3/2x)⁴

Using binomial expansion of ( a + b)ⁿ

here a = 2x/3 and b = -3/2x

n = 4

Expanding we get

⁴C₀(2x/3)⁴ + ⁴C₁(2x/3)³(-3/2x) +⁴C₂(2x/3)²(-3/2x)² + ⁴C₃(2x/3)(-3/2x)³

+ ⁴C₄(-3/2x)⁴

= 16/81*x⁴ - 4*8/27*3/2*x² + 6*4/9*9/4 - 4*2/3*27/8*1/x² + 81/16*x⁴

= 16x⁴/81 - 16x²/9 + 6 - 9/x² + 81/(16x⁴)

Hope, it helps !

Answer 2

Solution :

$\rm \Big(\frac{2x}{3} - \frac{3}{2x} \Big)^{4}$

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

= ( 2x/3)⁴-4×(2x/3)³(3/2x)

+ 6(2x/3)²(3/2x)²-4(2x/3)(3/2x)³+(3/2x)⁴

= 16x⁴/81 - 16x²/9 + 6 - 9/x²

+ 81/16x⁴

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