The condition that {x^3} - 3px + 2q may be di | vedclass

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(C) Given that ${x^2} + 2ax + {a^2} = {(x + a)^2}$ is a factor of $f(x) = {x^3} - 3px + 2q$.Since $(x+a)^2$ is a factor,$x = -a$ must be a root of $f(

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${p^3} = {q^2}$

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(C) Given that ${x^2} + 2ax + {a^2} = {(x + a)^2}$ is a factor of $f(x) = {x^3} - 3px + 2q$.Since $(x+a)^2$ is a factor,$x = -a$ must be a root of $f(x) = 0$ and $f'(x) = 0$.First,$f(-a) = {(-a)^3} - 3p(-a) + 2q = 0$.$-{a^3} + 3pa + 2q = 0$  ...$(i)$Next,$f'(x) = 3{x^2} - 3p$.Setting $f'(-a) = 0$,we get $3{(-a)^2} - 3p = 0$.$3{a^2} = 3p \Rightarrow p = {a^2}$ ...$(ii)$Substitute $p = {a^2}$ into equation $(i)$:$-{a^3} + 3({a^2})a + 2q = 0$$-{a^3} + 3{a^3} + 2q = 0$$2{a^3} + 2q = 0 \Rightarrow {a^3} = -q$.Squaring both sides,we get ${a^6} = {q^2}$.Since $p = {a^2}$,then ${p^3} = {({a^2})^3} = {a^6}$.Therefore,${p^3} = {q^2}$.

The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is

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