If a, b, c are real and x^3 - 3b^2x + 2c^3 is | vedclass

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(C) Let $f(x) = x^3 - 3b^2x + 2c^3$. Since $f(x)$ is divisible by $x - a$ and $x - b$,we have $f(a) = 0$ and $f(b) = 0$.$f(b) = b^3 - 3b^2(b) + 2c^3 =

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$a = b = c$ or $a = -2b = -2c$

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(C) Let $f(x) = x^3 - 3b^2x + 2c^3$. Since $f(x)$ is divisible by $x - a$ and $x - b$,we have $f(a) = 0$ and $f(b) = 0$.$f(b) = b^3 - 3b^2(b) + 2c^3 = b^3 - 3b^3 + 2c^3 = -2b^3 + 2c^3 = 0$.This implies $b^3 = c^3$,so $b = c$ (since $b, c$ are real).Now,$f(a) = a^3 - 3b^2a + 2c^3 = 0$. Substituting $c = b$,we get $a^3 - 3ab^2 + 2b^3 = 0$.Factoring the expression: $(a - b)(a^2 + ab - 2b^2) = 0$.$(a - b)(a - b)(a + 2b) = 0$.This gives $a = b$ or $a = -2b$.Since $b = c$,the solutions are $a = b = c$ or $a = -2b = -2c$.

If $a, b, c$ are real and $x^3 - 3b^2x + 2c^3$ is divisible by $x - a$ and $x - b$,then:

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