Suppose $x_1,x_2$ and $x_3$ are roots of $(11 - x)^3 + (13 - x)^3 - (24 - 2x)^3$ . What is the sum of $x_1 + x_2 + x_3$?

Question

Suppose $x_1,x_2$ and $x_3$ are roots of $(11 - x)^3 + (13 - x)^3 - (24 - 2x)^3$ . What is the sum of $x_1 + x_2 + x_3$ ?

What I Tried :- I expanded the expression and got :- $$ \rightarrow (11^3 + 13^3 + 24^3) - 10x^3 - 33x(11 - x) - 39x(13 - x) - 144x(24 - 2x)$$ $$ \rightarrow 17352 - 10x^3 + (33 + 39 + 288)x^2 - (363 + 507 + 3456)x $$ $$ \rightarrow -10x^3 + 360x^2 - 4326x + 17352 $$

Now on Wolfram Alpha I get that this can be factorised as $(-2)(x - 12)(5x^2 - 120x + 723)$ , hence these are the $x_1,x_2,x_3$. So $x_1 + x_2 + x_3 = 5x^2 - (120 - 1)x + (723 - 12 - 2)$ .

$\rightarrow 5x^2 - 119x + 713$ .

I hope this is the final answer , and I went through a lot of calculations, so this method of getting the answer is definitely not that simple . I am seeking a far more easier way or a shortcut to do this problem , can anyone help ?

Answer 1

Hint:

Writing expression $ax^3+bx^2+cx+d$ as: $$a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)$$ reveals that the coefficient of $x^{2}$ takes value: $$-a\left(x_{1}+x_{2}+x_{3}\right)$$

Answer 2

In general, Vieta's formulas for a polynomial $a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ of degree $n>0$ state that the sum of all roots is equal to $$x_1+\cdots+x_n=-\frac{a_{n-1}}{a_n}.$$ Hence we need the coefficients next to $x^3$ and $x^2$, which are 6 and -216, respectively (and not $-10$ and $360$ according to your expansion). Hence the sum is $$x_1+x_2+x_3=-\frac{-216}{6}=36.$$

Answer 3

Hint: $$a^3+b^3-(a+b)^3=-3ab(a+b)$$

Answer 4

First method (solve equation)

Using the formula:$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$ We have: $$(11-x)^3+(13-x)^3-(24-2x)^3=(11-x)^3+(13-x-24+2x)((13-x)^2+(13-x)(24-2x)+(24-2x)^2)=(11-x)^3-(11-x)(7x^2-172x+1057)=(11-x)((11-x)^2-7x^2+172x-1057)=(x-11)(6x^2-150x+936)=(x-11)(x-12)(x-13)$$

Then we have: $$x_1=11; x_2=12; x_3=13 \Rightarrow x_1+x_2+x_3=11+12+13=36$$

Second method (Vieta's formula)

General Vieta's formula looks like: $$\sum_{1\leq i_1<...<i_k\leq n}\left(\prod_{j=1}^{k} r_{i_j} \right)=(-1)^k \frac{a_{n-k}}{a_n}$$

Then, for cubic polynomial: $$a_3 x^3 + a_2 x^2 + a_1 x + a_0 = 0$$

The sum of roots is equal to: $$x_1+x_2+x_3=-\frac{a_2}{a_3}$$

That's because we interested only in coefficients near $x^3$ and $x^2$. $$(11-x)^3=-x^3+33x^2-...$$ $$(13-x)^3=-x^3+39x-...$$ $$(24-2x)^3=-8x^3+288x^2-...$$

Then: $$(11-x)^3+(13-x)^3-(24-2x)^3=6x^3-216x^3-...\Rightarrow a_3=6;a_2=-216.$$ $$x_1+x_2+x_3=-\frac{a_2}{a_3}=\frac{216}{6}=36$$