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Chapter 2: Problem 134

Show that \((99+70 \sqrt{2})^{1 / 3}=3+2 \sqrt{2}\).

Short Answer

Expert verified
The short answer is: We showed that taking the cube root of the expression \((99+70\sqrt{2})\) is equal to \(3+2\sqrt{2}\) by first cubing the given result \((3+2\sqrt{2})^3\), then simplifying and combining the terms using the binomial formula. This led us to the expression \(99 + 70\sqrt{2}\), which matches the given expression, demonstrating that \((99+70\sqrt{2})^{1/3} = 3+2\sqrt{2}\).

Step by step solution

01

Cube the result.

First, we need to cube the given result \((3+2 \sqrt{2})^3\). Using the binomial formula \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\), where \(a = 3\) and \(b = 2 \sqrt{2}\), we get: \((3+2 \sqrt{2})^3 = 3^3 + 3(3^2)(2\sqrt{2}) + 3(3)(2\sqrt{2})^2 + (2\sqrt{2})^3\)
02

Simplify the expression.

Now, let's simplify the expression we got in Step 1: \(3^3 = 3 \times 3 \times 3 = 27\) \(3(3^2)(2\sqrt{2}) = 3 \times 9 \times 2\sqrt{2} = 54\sqrt{2}\) \(3(3)(2\sqrt{2})^2 = 3 \times 3 \times (2\sqrt{2} \times 2\sqrt{2}) = 9 \times 8 = 72\) \((2\sqrt{2})^3 = 2\sqrt{2} \times 2\sqrt{2} \times 2\sqrt{2} = 2\sqrt{2} \times 8 = 16\sqrt{2}\)
03

Combine the simplified terms.

Now that we have simplified the terms, we need to combine them: \((3+2\sqrt{2})^3 = 27 + 54\sqrt{2} + 72 + 16\sqrt{2} = 99 + 70\sqrt{2}\)
04

Compare the expressions.

In Step 3, we arrived at the expression \(99 + 70\sqrt{2}\). This is the same as the given expression \((99+70 \sqrt{2})^{1 / 3}\). Therefore, we have demonstrated that taking the cube root of \((99+70\sqrt{2})\) is equal to \(3+2\sqrt{2}\): \((99+70 \sqrt{2})^{1 / 3}=3+2 \sqrt{2}\)

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Most popular questions from this chapter

Suppose \(p\) and \(q\) are polynomials of degree 3 such that \(p(1)=q(1), p(2)=q(2), p(3)=q(3),\) and \(p(4)=q(4) .\) Explain why \(p=q\).

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