Simple note

If $x + y \ = \ 3$ and $x^{3} + y^{3} \ = \ 25$ , what is $x^{2} + y^{2}$

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

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

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 - 6 years ago

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 $x^2+y^2=(x+y)^2-2xy=3^2-2\times\frac{2}{9}=\boxed{\frac{77}{9}}$ .

Alex Li - 6 years ago

I think you can perhaps use $\rightarrow$ OR $\Rightarrow$ instead of => .

The $\LaTeX$ codes are \rightarrow and \Rightarrow respectively .

A Former Brilliant Member - 6 years ago

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

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

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

We resubstitute this value of $xy$ back to our first equation to get $x^2+y^2=\dfrac{77}{9}$

Prasun Biswas - 6 years ago

hm....What is the answer?

Frankie Fook - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$