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The condition that ${x^3} - 3px + 2q$ may be divisible by a factor of the form ${x^2} + 2ax + {a^2}$ is
$3p = 2q$
$3p + 2q = 0$
${p^3} = {q^2}$
$27{p^3} = 4{q^2}$
Solution
(c) Given that ${x^2} + 2ax + {a^2}$ is a factor of
${x^3} – 3px + 2q = 0$
Let ${x^3} – 3px + 2q = ({x^2} + 2ax + {a^2})(x + \lambda )$,
where $\lambda $ is a constant.
Then equating the coefficients of like powers of x on both sides,
${x^3} – 3px + 2q = {x^3} + (2a + \lambda ){x^2} + ({a^2} + 2a\lambda )x + \lambda {a^2}$
==> $2a + \lambda = 0$$ \Rightarrow \lambda = – 2a$ …..$(i)$
and $ – 3p = {a^2} + 2a\lambda $ …..$(ii)$
and $2q = \lambda {a^2}$ …..$(iii)$
Put the value of $\lambda $ in $(iii),$
==> $2q = – 2{a^3} \Rightarrow q = – {a^3}$ ..…$(iv)$
Put the value of $\lambda $ in $(ii),$
==>$ – 3p = {a^2} + 2a( – 2a) = {a^2} – 4{a^2} = – 3{a^2}$
==> $ – 3p = – 3{a^2} \Rightarrow p = {a^2}$
==>$p = {( – q)^{2/3}}$
==> ${p^3} = {q^2}$.
