How do i prove this?

If one root of the equation ax^2 + bx + c = 0 is square of the other, show that b^3 + a^2.c + a.c^2 = 3abc

2. If x,y are roots and one is the square of the other, then $\displaystyle x-y^2 =0$ or $\displaystyle x^2 - y = 0$. It follows that $\displaystyle 0=(x-y^2)(x^2-y)= x^3 +y^3 -xy -x^2y^2 = (x+y)^3 - 3xy^2-3x^2y -xy -x^2y^2 =$
$\displaystyle = (x+y)^3 -3xy(x+y) -xy -x^2y^2$
Now Viete's formulas say $\displaystyle x+y = -\frac{b}{a}$ and $\displaystyle xy=\frac{c}{a}$.
So $\displaystyle 0=-\frac{b^3}{a^3} -3\frac{c}{a}(-\frac{b}{a}) - \frac{c}{a} - \frac{c^2}{a^2} = -\frac{b^3}{a^3} +3\frac{cb}{a^2}- \frac{c}{a} - \frac{c^2}{a^2}$ from which by multipliyng by $\displaystyle a^3$ the conclusion follows.

3. Originally Posted by saberteeth
How do i prove this?

If one root of the equation ax^2 + bx + c = 0 is square of the other, show that b^3 + a^2.c + a.c^2 = 3abc
Or

$\displaystyle a{x^2} + bx + c = 0.$

$\displaystyle {x_1} = x_2^2 \Leftrightarrow \frac{{ - b + \sqrt {{b^2} - 4ac} }} {{2a}} = {\left( {\frac{{ - b - \sqrt {{b^2} - 4ac} }} {{2a}}} \right)^2} \Rightarrow$

$\displaystyle \Rightarrow - ab + a\sqrt {{b^2} - 4ac} = {b^2} + b\sqrt {{b^2} - 4ac} - 2ac \Leftrightarrow$

$\displaystyle \Leftrightarrow \left( {a - b} \right)\sqrt {{b^2} - 4ac} = {b^2} + ab - 2ac \Rightarrow$

$\displaystyle \Rightarrow \left( {{a^2} - 2ab + {b^2}} \right)\left( {{b^2} - 4ac} \right) = {b^4} + 2{b^2}\left( {ab - 2ac} \right) + {\left( {ab - 2ac} \right)^2} \Leftrightarrow$

$\displaystyle \Leftrightarrow {a^2}{b^2} - 4{a^3}c - 2a{b^3} + 8{a^2}bc + {b^4} - 4{b^2}ac =$

$\displaystyle = {b^4} + 2a{b^3} - 4a{b^2}c + {a^2}{b^2} - 4{a^2}bc + 4{a^2}{c^2} \Leftrightarrow$

$\displaystyle - 4{a^3}c - 2a{b^3} + 8{a^2}bc = 2a{b^3} - 4{a^2}bc + 4{a^2}{c^2} \Leftrightarrow$

$\displaystyle 4a{b^3} + 4{a^2}{c^2} + 4{a^3}c = 12{a^2}bc \Leftrightarrow$

$\displaystyle \Leftrightarrow {b^3} + {a^2}c + a{c^2} = 3abc.$