Question 850743
Question 850743
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if the sum of the roots of quadratic equation, ax^2+bx+c=0, is equal to the sum of cubes of their reciprocals then show that ab^2=3a^2c+c^3
<pre>
The sum of the roots of {{{ax^2+bx+c}}}{{{""=""}}}{{{"0"}}} is {{{-b/a}}}

Use the quadratic formula to find the sum of the cubes of the reciprocals of the roots:

{{{((2a)/(-b + sqrt( b^2-4ac )))^3 }}}{{{""+""}}}{{{((2a)/(-b - sqrt( b^2-4ac )))^3 }}}{{{""=""}}}{{{-b/a}}}

To make things easier, let {{{z}}}{{{""=""}}}{{{sqrt( b^2-4ac)}}}

{{{((2a)/(-b + z))^3 }}}{{{""+""}}}{{{((2a)/(-b - z))^3 }}}{{{""=""}}}{{{-b/a}}}

Combine fractions on the left

{{{(8a^3)/(-b + z)^3 }}}{{{""+""}}}{{{(8a^3)/(-b - z)^3 }}}{{{""=""}}}{{{-b/a}}}

Divide through by {{{8a^3}}}

{{{1/(-b + z)^3 }}}{{{""+""}}}{{{1/(-b - z)^3 }}}{{{""=""}}}{{{(-b)/(8a^4)}}}

{{{((-b + z)^3+(-b - z)^3)/((-b+z)^3(-b-z)^3) }}}{{{""=""}}}{{{(-b)/(8a^4)}}}

The numerator factors as the sum of two cubes:

{{{((-b+z)+(-b-z))((-b+z)^2-(-b+z)(-b-z)+(-b-z)^2)}}} 
{{{(-b+z-b-z)((b^2-2bz+z^2)-(b^2-z^2)+(b^2+2bz+z^2))}}}
{{{(-2b)(b^2-2bz+z^2-b^2+z^2+b^2+2bz+z^2)}}}
{{{(-2b)(3z^2+b^2)}}}

The denominator simplifies:

{{{(-b+z)^3(-b-z)^3}}}
{{{((-b+z)(-b-z))^3}}}
{{{(b^2-z^2)^3}}}

So we have

{{{((-2b)(3z^2+b^2))/(b^2-z^2)^3}}}

Since {{{z}}}{{{""=""}}}{{{sqrt( b^2-4ac)}}}
{{{z^2}}}{{{""=""}}}{{{ b^2-4ac}}}

Substitute for z²

{{{((-2b)(3(b^2-4ac)+b^2))/(b^2-(b^2-4ac))^3}}}{{{""=""}}}{{{(-b)/(8a^4)}}}

{{{((-2b)(3b^2-12ac+b^2))/(b^2-b^2+4ac)^3}}}{{{""=""}}}{{{(-b)/(8a^4)}}}

{{{((-2b)(4b^2-12ac))/(4ac)^3}}}{{{""=""}}}{{{(-b)/(8a^4)}}} 

{{{((-2b)(4)(b^2-3ac))/(4ac)^3}}}{{{""=""}}}{{{(-b)/(8a^4)}}}

{{{(-8b(b^2-3ac))/(64a^3c^3)}}}{{{""=""}}}{{{(-b)/(8a^4)}}}

Simplify the fraction on the left 

{{{(-b(b^2-3ac))/(8a^3c^3)}}}{{{""=""}}}{{{(-b)/(8a^4)}}}

Divide both sides by -b

{{{(b^2-3ac)/(8a^3c^3)}}}{{{""=""}}}{{{1/(8a^4)}}}

Multiply both sides by 8a³

{{{(b^2-3ac)/(c^3)}}}{{{""=""}}}{{{1/a}}}

Cross multiply:

{{{ab^2-3a^2c}}}{{{""=""}}}{{{c^3}}}

{{{ab^2}}}{{{""=""}}}{{{3a^2c+c^3}}}

Edwin</pre>