+0

# Tough Algebra Question, Completely Stuck, Please Help!

0
43
1

Let x and y be positive real numbers. Define  $$a = 1 + \frac{x}{y}$$  and $$b = 1 + \frac{y}{x}$$ . If  $$a^2 + b^2 = 15$$    compute $$a^3 + b^3$$

Apr 17, 2020

### 1+0 Answers

#1
+24933
+2

Let $$x$$ and $$y$$ be positive real numbers.
Define $$a = 1 + \dfrac{x}{y}$$ and  $$b = 1 + \dfrac{y}{x}$$.
If $$a^2 + b^2 = 15$$ compute $$a^3 + b^3$$

$$\begin{array}{|lrcll|} \hline (1) & a &=& 1 + \dfrac{x}{y} \\ & \mathbf{ \dfrac{x}{y} } &=& \mathbf{a-1} \\\\ (2) & b &=& 1 + \dfrac{y}{x} \\ & \dfrac{y}{x}&=& b-1 \\ &\mathbf{\dfrac{x}{y}} &=& \mathbf{\dfrac{1}{ b-1}} \\\\ \hline & \mathbf{\dfrac{x}{y} =} a-1 &=& \dfrac{1}{ b-1} \\ & a-1 &=& \dfrac{1}{ b-1} \\ & (a-1)(b-1) &=& 1 \\ & ab-(a+b)+1 &=& 1 \\ & \mathbf{ab} &=& \mathbf{a+b} \\ \hline & (ab)^2 &=& (a+b)^2 \\ & (ab)^2 &=& a^2+2ab+b^2 \\ & (ab)^2 &=& a^2+b^2+ 2ab \quad | \quad a^2 + b^2 = 15 \\ & (ab)^2 &=& 15+ 2ab \\ & \mathbf{(ab)^2 - 2ab -15} &=& \mathbf{ 0 } \\\\ & ab &=& \dfrac{2\pm\sqrt{4-4*(-15)}} {2} \\ & ab &=& \dfrac{2\pm\sqrt{4+4*15}} {2} \\ & ab &=& \dfrac{2\pm\sqrt{64}} {2} \\ & ab &=& \dfrac{2\pm 8} {2} \\ & ab &=& \dfrac{2 + 8} {2} \\ &&& \text{a and b be positive, while x and y be positive}, \\ &&& \text{so ab be positive } \\ & \mathbf{ab} &=& \mathbf{5} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a^3+b^3 &=& (a+b)(a^2-ab+b^2) \\ a^3+b^3 &=& (a+b)(a^2+b^2-ab) \quad & | \quad a^2+b^2 = 15 \\ a^3+b^3 &=& (a+b)(15-ab) \quad & | \quad a+b=ab=5 \\ a^3+b^3 &=& 5(15-5) \\ a^3+b^3 &=& 5*10 \\ \mathbf{a^3+b^3} &=& \mathbf{50} \\ \hline \end{array}$$

Apr 17, 2020