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If (x + ½)²=3. , then what is x3 +1/x3 ?

A1

B√3

C0

D3√3

Answer:

C. 0

Read Explanation:

If (x+1x)2=3(x + \frac{1}{x})^2 = 3, then the value of x3+1x3x^3 + \frac{1}{x^3} is 0.


If (x+1x)2=3(x + \frac{1}{x})^2 = 3, then x+1x=3x + \frac{1}{x} = \sqrt{3}.
Whenever x+1x=3x + \frac{1}{x} = \sqrt{3}, the value of x3+1x3x^3 + \frac{1}{x^3} is always 0 (and x6=1x^6 = -1).


Step 1: Find the value of x+1xx + \frac{1}{x}
(x+1x)2=3(x + \frac{1}{x})^2 = 3
Taking the square root on both sides:
x+1x=3x + \frac{1}{x} = \sqrt{3}

Step 2: Use the algebraic identity for cubes
We know the identity:
x3+1x3=(x+1x)33(x+1x)x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x})

Step 3: Substitute the value
Substitute (x+1x)=3(x + \frac{1}{x}) = \sqrt{3} into the identity:
x3+1x3=(3)33(3)x^3 + \frac{1}{x^3} = (\sqrt{3})^3 - 3(\sqrt{3})
x3+1x3=3333x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3}
x3+1x3=0x^3 + \frac{1}{x^3} = 0


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