#### Question

If the roots of the equation (c^{2} – ab) x^{2} – 2 (a^{2} – bc) x + b^{2} – ac = 0 in x are equal, then show that either a = 0 or a^{3} + b^{3} + c^{3} = 3abc

#### Solution

The given quadric equation is `(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0`

THen prove that either a = 0 or `a^3 + b^3 + c^3 = 3abc`

Here

`a = (c^2 - ab), b = -2(a^2 - bc) " and " c = (b^2 - ac)`

As we know that `D = b^2 - 4ac`

Putting the value of `a = (c^2 - ab), b = -2(a^2 - bc) and c = (b^2 - ac)`

The given equation will have real roots, if D = 0

`4a(a^3 + b^3 + c^3 - 3abc) = 0`

`a(a^3 + b^3 + c^3 - 3abc) = 0`

So , either

a = 0

or

`(a^3 + b^3 + c^3 - 3abc) = 0`

`a^3 + b^3 + c^3 = 3abc`

Hence, a = 0 or `a^3 + b^3 + c^3 = 3abc`

Is there an error in this question or solution?

Solution If the Roots of the Equation (C2 – Ab) X2 – 2 (A2 – Bc) X + B2 – Ac = 0 in X Are Equal, Then Show that Either a = 0 Or A3 + B3 + C3 = 3abc Concept: Nature of Roots.