Question

If the roots of the quadratic equation (62 - ab)x² - 2 (02 - bc)x + b2 - ac = 0) in x are equal, then show
that either a = 0 or a + b3 + c3 = 3abc​

Answer 1

Answer:

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Answer 2

$\huge\tt\blue{Answer:-}$

Let, $p {x}^{2} + qx + r = 0$ be a quadratic polynomial.

Since, roots of the quadratic equation are equal so, ${q}^{2} - 4pr = 0$

Here,

$q = - 2( {a}^{2} - bc)$

$p = {c}^{2} - ab$

$r = {b}^{2} - ac$

Since,

${q}^{2} - 4pr = 0 \\ = > {q}^{2} = 4pr \\ = > {( - 2( {a }^{2 } - bc)) }^{2} = 4 \times( {c}^{2} - ab)( {b}^{2} - ac) \\ = > 4( {a}^{4} + {b}^{2} {c}^{2} - 2 {a}^{2} bc) = 4( {b}^{2} {c}^{2} + {a}^{2} bc - a {b}^{3} - {ac}^{3}) \\ = > {a}^{4} + {b}^{2} {c}^{2} - 2 {a}^{2} bc = {b}^{2} {c}^{2} + {a}^{2} bc - {ab}^{3} - {ac}^{3} \\ = > {a}^{4} + {b}^{2} {c}^{2} - 2 {a}^{2} bc - {b}^{2} {c}^{2} - {a}^{2} bc + {ab}^{3} + {ac}^{3} = 0 \\ = > {a}^{4} + {ab}^{3} + {ac}^{3} - 3 {a}^{2} bc = 0 \\ = > a( {a}^{3} + {b}^{3} + {c}^{3} - 3 abc ) = 0\\ \\ so \\ a = 0 \: \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: \: \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

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