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

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Question

(c) As $f(x) = {x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and $x - b$, therefore
$f(a) = 0\,\,$

$\Rightarrow {a^3} - 3{b^2}a + 2{c^3} = 0$.

Vedclass · Accepted Answer

$a = b = c$,$a = - 2b = - 2c$

Vedclass · Answer

(c) As $f(x) = {x^3} - 3{b^2}x + 2{c^3}$ is divisible by $x - a$ and $x - b$, therefore
$f(a) = 0\,\,$

$\Rightarrow {a^3} - 3{b^2}a + 2{c^3} = 0$.....$(i)$

and $f(b) = 0$==>${b^3} - 3{b^3} + 2{c^3} = 0$.....$(ii)$

From $(ii),$ $b = c$

From $(i)$, ${a^3} - 3a{b^2} + 2{b^3} = 0$ (Putting $b = c$)

$ \Rightarrow (a - b)({a^2} + ab - 2{b^2}) = 0$

$ \Rightarrow $$a = b$ or ${a^2} + ab = 2b$

Thus $a = b = c$ or ${a^2} + ab = 2{b^2}$ and $b = c$

${a^2} + ab = 2{b^2}$is satisfied by $a = - 2b$. But $b = c$.

$	herefore$ ${a^2} + ab - 2{b^2}$ and $b = c$ is equivalent to $a = - 2b = - 2c$

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

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