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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
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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 \(a^3 + b^3\)

 

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

 

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

 

laugh

 Apr 17, 2020

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