Question - Simplified Cube Root of a Fraction
Solution:
The expression in the image is:$$\left(\frac{2 + 2\sqrt{13}}{6}\right)^3$$To solve this expression, you need to raise the fraction within the parentheses to the power of 3. First, simplify the fraction by dividing both the numerator and the denominator by 2, then raise the result to the third power:Simplify the fraction:$$\left(\frac{2}{6} + \frac{2\sqrt{13}}{6}\right)^3 = \left(\frac{1}{3} + \frac{\sqrt{13}}{3}\right)^3$$Now compute the cube of the simplified expression. Let's denote $$a = \frac{1}{3}$$ and $$b = \frac{\sqrt{13}}{3}$$, then the expression becomes $$(a + b)^3$$. Using the binomial expansion:$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$Apply this formula:$$\left(\frac{1}{3} + \frac{\sqrt{13}}{3}\right)^3 = \left(\frac{1}{3}\right)^3 + 3\left(\frac{1}{3}\right)^2\left(\frac{\sqrt{13}}{3}\right) + 3\left(\frac{1}{3}\right)\left(\frac{\sqrt{13}}{3}\right)^2 + \left(\frac{\sqrt{13}}{3}\right)^3$$Now carry out the exponentiation and multiplication:$$= \frac{1}{27} + 3 \cdot \frac{1}{9} \cdot \frac{\sqrt{13}}{3} + 3 \cdot \frac{1}{3} \cdot \frac{13}{9} + \frac{13\sqrt{13}}{27}$$Simplify the terms:$$= \frac{1}{27} + \frac{\sqrt{13}}{9} + \frac{13}{9} + \frac{13\sqrt{13}}{27}$$Combine the terms:$$= \frac{1 + 3\sqrt{13} + 39 + 13\sqrt{13}}{27}$$Combine like terms:$$= \frac{40 + 16\sqrt{13}}{27}$$That's the simplified form of the cube of the original expression:$$\left(\frac{2 + 2\sqrt{13}}{6}\right)^3 = \frac{40 + 16\sqrt{13}}{27}$$
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