If $x^2 + y^2 = 1$, then find the value of $(3x-4x^3)^2+(3y-4y^3)^2$ where $x$, $y \in \mathbb {R}$ .

What I've tried :

I thought $x$, $y$ as $\sin \theta$ and $\cos \theta$ .

But, it doesn't seems to work and I'm unable to see any particular pattern in it .

  • 5
    $\begingroup$ $(x+i y)^3+3(x-i y)(x^2+y^2-1)=(-3x+4x^3)+i(3y-4y^3)$ $\endgroup$ Apr 18 at 11:02
  • $\begingroup$ @IvanKaznacheyeu I've added an answer based on this beautiful identity $\endgroup$
    – user
    Apr 18 at 16:55

6 Answers 6


Your first idea was the right one

Hint. Try to express $\cos(3\theta)$ and $\sin(3\theta)$ as functions of $\cos(\theta)$ and $\sin(\theta)$ respectively, and see what happenes.

  • 1
    $\begingroup$ Even easier with $6\theta$. $\endgroup$ Apr 18 at 9:09
  • $\begingroup$ too much simple! $\endgroup$
    – user
    Apr 18 at 9:13

You do not need to use $\sin$ or $\cos$.

Just rearrange the expression and use $$1=(x^2+y^2)^2 = x^4+y^4+2x^2y^2$$ $$1= (x^2+y^2)^3 = x^6+y^6 + 3 x^2 y^2 (x^2+y^2) = x^6+y^6 + 3x^2 y^2$$ So you get $$(3x-4x^3)^2+(3y-4y^3)^2 = x^2(3-4x^2)^2 + y^2(3-4y^2)^2$$ Expanding and using $x^2+y^2=1$ gives $$= 9-24(x^4+y^4)+16(x^6+y^6)= 9-24(1-2x^2y^2)+16(1-3x^2y^2)$$ $$= 9-24+16 = \boxed{1}$$


An answer that doesn't use algebraic expansion :


$$\begin{align} P(x,y)&:=(3x-4x^3)^2+(3y-4y^3)^2\end{align}$$

Making the substitution $u=x^2, \thinspace u\in[0,1]$ and $y^2=1-u$, then we can observe that the polynomial $P(x,y)$ is equivalent to the polynomial

$$ \begin{align}Q(u):=u(3-4u)^2&+(1-u)(4u-1)^2\thinspace .\end{align} $$

Then using the quick property $Q(u)=Q(1-u),\thinspace \forall u\in[0,1]$ we see that : If $u\in\left\{0,1,\frac 14,\frac 34\right\}$, then $Q(u)=1\thinspace .$

Finally we observe that, if $0<\deg \{Q\}≤3\thinspace ,$ then the polynomial $Q(u)-1$ has at most $3$ real roots, which leads to a contradiction .

Therefore, we conclude that :

$Q(u)\equiv 1\thinspace,$ since $\deg\left\{Q\right\}=0\thinspace .$ This means, if $(x,y)\in\mathbb R^{2}$ and $x^2+y^2=1$, then $P(x,y)=1\thinspace .$

  • 2
    $\begingroup$ That works out nicely (+1). Just a nitpick on the "therefore" part, you proved that $Q(u)-1 \equiv 0$ as a polynomial, so you have $Q(u)=1$ on the whole $\mathbb R$ not just on $[0,1]$. You could have just as well chosen $\{0, 1, -1, 2\}$ as values to check that $Q(u)=1$, and the conclusion would have been the same, since at that point you are looking at the polynomial $Q(u)$ even if the $u$ of interest is restricted to $[0,1]$. $\endgroup$
    – dxiv
    Apr 18 at 18:49
  • 1
    $\begingroup$ @dxiv I think I got your points. Thank you very much! $\endgroup$ Apr 18 at 21:08

Another way following the fine suggestion given by Ivan Kaznacheyeu in the comments

$$x^2+y^2=1 \iff z=x+iy = e^{i\theta}$$

and since in general

$$(-3x+4x^3)+i(3y-4y^3)= (x+i y)^3+3(x-i y)(x^2+y^2-1)$$

by the given condition we have

$$w=(-3x+4x^3)+i(3y-4y^3)= (x+i y)^3=z^3=e^{i3\theta}$$

that is



Another "long" way by

  • $x=\frac{2t}{1+t^2}$
  • $y=\frac{1-t^2}{1+t^2}$

such that $x^2+y^2=1$ and then




$$=\frac{t^{12} + 6 t^{10} + 15 t^{8} + 20 t^6 + 15 t^4 + 6 t^2 + 1}{(1+t^2)^6}=\frac{(1+t^2)^6}{(1+t^2)^6}=1$$


Let $x^2=u+{1\over 2},$ $y^2=v+{1\over 2}.$ Then $u+v=0.$ The quantity in question takes the form $P(u)+P(-u),$ where $$P(u)=\left (u+{1\over 2}\right )(1-4u)^2$$ The problem becomes whether the quantity $P(u)+P(-u)$ is constant. It is equivalent to $$P'(u)-P'(-u)=0$$ i.e. the function $P'(u)$ is even. We have $$P'(u)=(4u-1)^2+8\left (u+{1\over 2}\right )(4u-1)\\ =(4u-1)(4u-1+8u+4)=3(16u^2-1)$$ Hence $P'(u)$ is indeed an even polynomial. Once we know that $P(u)+P(-u)$ is independent of $u$ we plug in $u=0$ and get $$P(u)+P(-u)=2P(0)=1$$


  1. The advantage of this method is that we are dealing with expressions of the second degree. Moreover the method can be applied to other expressions of similar form.
  2. If we are too lazy to verify whether $P'$ is even, we can instead verify if $P''$ is odd. Then we are dealing with a linear function. Once we know that $P''$ is odd the polynomial $P'$ is even.

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