Question

if the sum of the roots of the quadratic equation
$a {x}^{2} + \: bx \: + \: c = \: 0$
is equal to the sum of the cubes of the reciprocals, then prove that
$a {b}^{2} = 3 {a}^{2}c \: + \: {c}^{3}$
​

Answer 1

Step-by-step explanation:

let roots of the equation be p and q

given condition sum of roots is equal to the sum of the cubes of the reciprocals

$p + q = (1 \div {p}^{3} ) + (1 \div {q}^{3} )$

then

$(p + q) = ( {p}^{3} + {q}^{3} ) \div {(pq)}^{3} \\ (p + q) = (p + q)( {p}^{2} + {q}^{2} - pq) \div {(pq)}^{3} \\ 1 = ( {p}^{2} + {q}^{2} - pq) \div {(pq)}^{3} \\ {(pq)}^{3} = {p}^{2} + {q}^{2} - pq \\ {(pq)}^{3} + pq = {p}^{2} + {q}^{2} \\ (pq)^{3} + pq= {(p + q)}^{2} - 2pq \\ {(pq)}^{3} + 3pq = {(p + q)}^{2}$

here pq=(c/a) and p+q=(-b/a)

substitute the values in above equation

${(c \div a)}^{3} + (c \div a) = {( - a \div b)}^{2} \\ ( {c}^{3} \div {a}^{3} ) + (c \div a) = {a}^{2} \div {b}^{2} \\( {c}^{3} + {a}^{2}) \div {a}^{3} = {a}^{2} \div {b}^{2} \\ {b}^{2}( {c}^{3} + {a}^{2} ) = {a}^{2} ( {a}^{3} ) \\ {b}^{2} ( {c}^{3} + {a}^{2} ) = {a}^{5}$

Answer 2

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$to \: prove \: that : \\ ab {}^{2} = 3a {}^{2} c + c {}^{3} \\ \\ given \: quadratic \: equation \: is \\ ax {}^{2} + bx + c = 0 \\ let \: its \: roots \: be \: \alpha \: and \: \beta \\ \\ so \: w e\: know \: that \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha \beta = \frac{c}{a} \\ \\ moreover \:we \: have \\ \alpha {}^{3} + \beta {}^{3} = ( \alpha + \beta ) {}^{3} - 3 \alpha \beta ( \alpha + \beta ) \\ \\ = ( \frac{ -b \: }{a} ) {}^{3} - 3( - \frac{b}{a} \: )( \frac{c}{a} \: ) \\ \\ = \frac{ - b {}^{3} }{a {}^{3} } + \frac{3bc}{a {}^{2} } \\ \\ = \frac{3abc -b {}^{3} }{a {}^{3} }$

$according \: to \: given \: condition \\ \alpha + \beta = \frac{1}{ \alpha {}^{3} } + \frac{1}{ \beta {}^{3} } \\ \\ \alpha + \beta = \frac{ \alpha {}^{3} + \beta {}^{3} }{ \alpha {}^{3}. \beta {}^{3} } \\ \\( \alpha + \beta )( \alpha \beta ) {}^{3} = \alpha {}^{3} + \beta {}^{3} \\ \\ (- \frac{b}{a} \: )( \frac{c}{a} ) {}^{3} = \frac{3abc - b {}^{3} }{a {}^{3} } \\ \\ \frac{ - bc {}^{3} }{a {}^{4} } = \frac{3abc - b {}^{3} }{a {}^{3} } \\ \\ - \frac{bc {}^{3} }{a} = 3abc - b {}^{3} \\ \\ - bc {}^{3} = b(3a {}^{2} c - b {}^{2} ) \\ \\ - c {}^{3} = 3a {}^{2} c - b {}^{2} \\ \\ b {}^{2} = 3a {}^{2} c + c {}^{3}$