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Given that $x,y$ real numbers such that :

$x^2+xy+y^2=4$

And

$x^4+x^2y^2+y^4=8$

How can one calculate :

$x^6+x^3y^3+y^6$

Can someone give me hint .

Ivan Neretin
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user233658
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3 Answers3

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Note that $$(x^2+xy+y^2)^2-2xy(x^2+xy+y^2)=x^4+x^2y^2+y^4 $$ so that $$ xy = 1.$$ Then $$\begin{align}x^6+x^3y^3+y^6&=(x^4+x^2y^2+y^4)(x+xy+y^2)-xy^5-2x^2y^4-2x^4y^2-yx^5\\ &=8\cdot 4-xy(x^4-2x^3y-2xy^3-y^4)\\ &=32-1\cdot ((x^4+x^2y^2+y^4)-xy(2x^2-xy-2y^2))\\ &=24+1\cdot(2(x^2+xy+y^2)-3xy)\\ &=29 \end{align} $$ (I suppose) Just keep subtracting simple products and powers of the given polynomials to get rid of moniomials not divisible by $xy$, then divide out $xy$ from the rest and continue. Alternatively to the above, you might also start by subtracting $(x^2+xy+y^2)^3$, which would also "kill" the $x^6$ and $y^6$.

Hagen von Eitzen
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    small typo: $$x^6+x^3y^3+y^6=...=8\cdot 4-xy(x^4+2x^3y+2xy^3+y^4)$$ $$=32-1\cdot \left( (x^4+x^2y^2+y^4)+xy(2x^2-xy+2y^2)\right)$$ $$=24-1\cdot(2(x^2+xy+y^2)-3xy)=24-(8-3)=19.$$ – Oleg567 Oct 22 '15 at 21:59
  • Yes, I used an entirely different (and much stupider) method to get $19$ as well, and just found Hagen’s slip. – Lubin Oct 22 '15 at 22:31
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why not. $$ x^4 + x^2 y^2 + y^4 = (x^2 + xy + y^2)(x^2 - xy+y^2). $$ Worth memorizing. So $$ x^2 - xy + y^2 = 2. $$ Also $$ 2xy = 2, \; \; \; xy=1. $$ And $$ x^2 + y^2 = 3. $$ $$ 27 = (x^2 + y^2)^3 = x^6 + 3 x^4 y^2 + 3 x^2 y^4 + y^6 = x^6 + y^6 + 3 x^2 y^2 (x^2 + y^2) = x^6 + y^6 + 3 \cdot 1 \cdot 3 = x^6 + y^6 + 9 $$ So, $$x^3 y^3 = 1, \; \; \; x^6 + y^6 = 18, $$ $$ x^6 + x^3 y^3 + y^6 = 19. $$

Will Jagy
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  • I like this one best of all – Lubin Oct 22 '15 at 22:48
  • @Lubin, thanks. – Will Jagy Oct 22 '15 at 22:52
  • You should have seen the monstrosity that I created to get the right answer. Rotated the figures by $45^\circ$ to get equations in $X^2$ and $Y^2$ only, solved the first for $X^2$, found $x=\phi$ (Golden Ratio), $y=1/\phi$, and calculated the value of the sextic there. – Lubin Oct 23 '15 at 02:17
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Denote $a=x/y$, $b=xy$.

Then $$a+1+\frac{1}{a}=\frac{4}{b},$$ $$a^2+1+\frac{1}{a^2}=\frac{8}{b^2},$$ $$a^3+1+\frac{1}{a^3}=?$$

Then from the $1$st equation: $$ \left(a+\frac{1}{a}\right)^2 = \left(\frac{4}{b}-1\right)^2, $$ combining it with $2$nd equation, $\Rightarrow$ $b=1$.

Then, $a+\frac{1}{a}=3$, $a^2+\frac{1}{a^2}=7$; therefore $a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}\right) - \left(a+\frac{1}{a}\right) = 21-3=18$.

Then $a^3+1+\frac{1}{a^3}=19$. Myltiplying by $b^3=1$, $\Rightarrow$ $x^6+x^3y^3+y^6=19$.

Oleg567
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