# [SOLVED]Find a cubic function

#### anemone

##### MHB POTW Director
Staff member
Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.

• ##### Well-known member
Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)
So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$
Or $(\sqrt x)(x + 9) = - 9(x+1)$
Or squaring $x(x+9)^2 = 81(x+1)^2$
Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$
Or $x^3 - 63x^2 - 81 x - 81 = 0$
This is the required equation

• lfdahl, solakis and anemone

#### solakis

##### Active member
Find a polynomial of degree 3 with real coefficients such that each of its roots is equal to the square of one root of the polynomial $P(x)=x^3+9x^2+9x+9$.
Does the above mean that if a,b,c are the 3 roots of the above polynomial then the the roots of the new polynomial must be $a^2 ,b^2 , c^2$ ??

• anemone

#### solakis

##### Active member
Because the square of the roots shall be roots of the required polynomial so we must have $\sqrt x$ as roots of P(X)
So $(\sqrt x)^3 + 9 x + 9 (\sqrt x) + 9 = 0$
Or $(\sqrt x)(x + 9) = - 9(x+1)$
Or squaring $x(x+9)^2 = 81(x+1)^2$
Or $x(x^2 + 18 x + 81) = 81 x^2 + 162x + 81$
Or $x^3 - 63x^2 - 81 x - 81 = 0$
This is the required equation
kaliprasand must we not have also $-\sqrt x$ .Anyway i did this problem in a different way and i got the same answer

• ##### Well-known member
kaliprasand must we not have also $-\sqrt x$ .Anyway i did this problem in a different way and i got the same answer
you are right and it is my mistake

as $\sqrt{x}$ is positive by definition so my line should be $\pm \sqrt{x}$ and the we get the same result because of squaring

• solakis and anemone

#### solakis

##### Active member

Let a,b,c be the roots of P(x) = $x^3+9x^2+9x+9$
And $a^2, b^2 ,c^2$ the roots of P(y)= $y^3+Ay^2+By+C$
From the polynomial properties we have:
P(x) = $x^3+9x^2+9x+9$=(x-a)(x-b)(x-c) $\Rightarrow$...............
.a+b+c=-9................... (1)
ab+ac+bc=9 ................(2)
abc=-9 ................................(3)
And also P(y)= $y^3+Ay^2+By+C$=$(y-a^2)(y-b^2)(y-c^2)\Rightarrow$
$a^2+b^2+c^2=-A$.................................................(4)
$a^2b^2 +a^2c^2+b^2c^2=B$...............................(5)
$a^2b^2c^2=-C$.............................................................(6)
From(3) we have:$a^2b^2c^2=81$ hence
C=-81.....................................................................................(7)
From (1) and squaring we get :
$(a+b+c)^2=81\Rightarrow a^2+b^2+c^2= 81-2(ab+ac+bc)$ and using (2) we get :
$a^2+b^2+c^2= 63$ and hence :
A= -63.................................................................................(8)
From (2) and squaring we get:
$(ab+ac+bc)^2=81\Rightarrow a^2b^2+a^2c^2+b^2c^2+2(a^2bc+ab^2c+abc^2)=81\Rightarrow a^2b^2+a^2c^2+b^2c^2+2(a(abc)+b(abc)+c(abc))=81$ and using (3) we have:
$a^2b^2+a^2c^2+b^2c^2 =81-2(-9a-9b-9c)=81+18(a+b+c)$ and using (1) we have:
$a^2b^2+a^2c^2+b^2c^2=81+18(-9)=-81$ hence:
B=-81.................................................................................(9)
And P(y)= $y^3+Ay^2+By+C =y^3-63y^2-81y-81$ by using (7),(8),(9)

Last edited by a moderator:
• 