Find the cube of:
x−12x -\dfrac{1}{2}x−21
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x−12x - \dfrac{1}{2}x−21
Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
=(x)3−(12)3−3×x2×(12)+3×x×(12)2=x3−(18)−(3x22)+(3x4)= (x)^3 - \Big(\dfrac{1}{2}\Big)^3 - 3 \times x^2 \times \Big(\dfrac{1}{2}\Big) + 3 \times x \times \Big(\dfrac{1}{2}\Big)^2\\[1em] = x^3 - \Big(\dfrac{1}{8}\Big) - \Big(\dfrac{3x^2}{2}\Big) + \Big(\dfrac{3x}{4}\Big)=(x)3−(21)3−3×x2×(21)+3×x×(21)2=x3−(81)−(23x2)+(43x)
Hence, x+12=x3−(18)−(3x22)+(3x4)x + \dfrac{1}{2} = x^3 - \Big(\dfrac{1}{8}\Big) - \Big(\dfrac{3x^2}{2}\Big) + \Big(\dfrac{3x}{4}\Big)x+21=x3−(81)−(23x2)+(43x)
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3b - 2a
2x+1x2x +\dfrac{1}{x}2x+x1
If x−1x=3x -\dfrac{1}{x}= 3x−x1=3, the value of x2+1x2x^2 +\dfrac{1}{x^2}x2+x21 is:
0
7
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If a + b = 7 and ab = 10; the value of a2 + b2 is equal to:
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49
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