Cubing the first equation, \(x^3+3x^2y+3xy^2+y^3=27=>3xy(x+y)=27-(x^3+y^3)=2\), so \(xy=\frac{2}{9}\). Now, note that \(x^2+y^2=(x+y)^2-2xy=3^2-2\times\frac{2}{9}=\boxed{\frac{77}{9}}\).
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Alex Li
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1 year, 7 months ago

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@Alex Li
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Just for the sake of mentioning, one can overkill this problem using Newton's Identities:

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TopNewestCubing the first equation, \(x^3+3x^2y+3xy^2+y^3=27=>3xy(x+y)=27-(x^3+y^3)=2\), so \(xy=\frac{2}{9}\). Now, note that \(x^2+y^2=(x+y)^2-2xy=3^2-2\times\frac{2}{9}=\boxed{\frac{77}{9}}\). – Alex Li · 1 year, 7 months ago

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\[x^2+y^2=(x+y)^2-2xy\implies x^2+y^2=9-2xy\]

\[x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)\implies 25=3(x^2+y^2-xy)\implies 25=3(9-2xy-xy)\implies xy=\frac 29\]

We resubstitute this value of \(xy\) back to our first equation to get \(x^2+y^2=\dfrac{77}{9}\) – Prasun Biswas · 1 year, 7 months ago

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The \(\LaTeX\) codes are

`\rightarrow`

and`\Rightarrow`

respectively . – Azhaghu Roopesh M · 1 year, 7 months agoLog in to reply

\[x^3+y^3 = (x+y)(x^2-xy+y^2) \\\Rightarrow 25 = 3(x^2-xy+y^2) \\\Rightarrow x^2-xy+y^2=\dfrac{25}{3} \dots (1) \\\Rightarrow (x+y)^2=3^2 \\\Rightarrow x^2+2xy+y^2=9 \\ \Rightarrow \dfrac{x^2}{2} + xy + \dfrac{y^2}{2} = \dfrac{9}{2} \dots (2) \]

Adding \((1),(2)\) , we have:

\[\dfrac{3x^2}{2}+\dfrac{3y^2}{2}= \dfrac{25}{3}+\dfrac{9}{2} \\\Rightarrow \dfrac{3}{2}(x^2+y^2)= \dfrac{77}{6} \\ \Rightarrow \boxed{x^2+y^2= \dfrac{77}{9}} \] – Nihar Mahajan · 1 year, 7 months ago

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hm....What is the answer? – Frankie Fook · 1 year, 7 months ago

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