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

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(C) Given that ${x^2} + 2ax + {a^2}$ is a factor of ${x^3} - 3px + 2q$.Note that ${x^2} + 2ax + {a^2} = {(x + a)^2}$.Let ${x^3} - 3px + 2q = {(x + a)^

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

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(C) Given that ${x^2} + 2ax + {a^2}$ is a factor of ${x^3} - 3px + 2q$.Note that ${x^2} + 2ax + {a^2} = {(x + a)^2}$.Let ${x^3} - 3px + 2q = {(x + a)^2}(x - 2a) = (x^2 + 2ax + a^2)(x - 2a)$.Expanding the right side: $(x^2 + 2ax + a^2)(x - 2a) = x^3 - 2ax^2 + 2ax^2 - 4a^2x + a^2x - 2a^3 = x^3 - 3a^2x - 2a^3$.Comparing this with ${x^3} - 3px + 2q$,we get:$-3p = -3a^2 \Rightarrow p = a^2$.$2q = -2a^3 \Rightarrow q = -a^3$.Now,substitute $a^2 = p$ into $q = -a^3$:$q^2 = (-a^3)^2 = a^6 = (a^2)^3 = p^3$.Thus,the condition is ${p^3} = {q^2}$.

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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