# Question #ec0d2

##### 3 Answers

See the **Proof** given in the **Explanation Section** below.

#### Explanation:

Let the **Roots** of the given **Quadr.** be

By what has been given, we may, assume that,

It is also known that,

Hence,

With

This completes the **Proof.**

**Enjoy Maths.!**

See below.

#### Explanation:

if

substituting into

we obtain the polynomial

and for

with the respective roots

and as we can verify

or

concluding:

The assertion is true. If

concerning the roots of

Suppose the roots of the general quadratic equation:

# ax^2+bx+c = 0 #

are

# "sum of roots" \ \ \ \ \ \= alpha+beta = -b/a #

# "product of roots" = alpha beta \ \ \ \ = c /a #

If one root is the square of the other than we can denote the roots by

Using the sum of the roots property:

# alpha + beta = -b/a => alpha + alpha^2 = -b/a#

# :. b = -a alpha(1 + alpha) .... [1]#

And the product of the rots property:

# alpha beta = c/a => alpha*alpha^2 = c/a => alpha^3 = c/a#

# :. c = a alpha^3 .... [2]#

Now, consider the LHS of the given identity, and substitute for

# LHS = b^3+ac(a+c) #

# " " = (-a alpha(1 + alpha))^3 + a(a alpha^3)(a+a alpha^3)#

# " " = -a^3 alpha^3(1 + alpha)^3 + a(a alpha^3)(a)(1 +alpha^3)#

# " " = -a^3 alpha^3(1 + alpha)^3 + a^3 alpha^3(1 +alpha^3)#

# " " = -a^3 alpha^3{(1 + alpha)^3-(1 +alpha^3)}#

# " " = -a^3 alpha^3{(1 +3 alpha + 3 alpha^2 + alpha^3)-(1 +alpha^3 )}#

# " " = -a^3 alpha^3(3 alpha + 3 alpha^2)#

# " " = -3 a^3 alpha^3(alpha + alpha^2)#

But we showed earlier that

Therefore.

# LHS = -3a^3(c/a)(-b/a)#

# " " = (3a^3bc)/a^2#

# " " = 3abc \ \ \ # QED