Question

If 'a' and 'b' are distinct zeroes of the polynomial $x^3-2x+c$ and $a^2(2a^2+4ab+3b^2)=3$ then $b^2(3a^2+4ab+2b^2)$ is equal to :

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (A)

3

Answer 2

The polynomial is $P(x) = x^3 - 2x + c$.

Since 'a' and 'b' are distinct zeroes of $P(x)$, we have:

$a^3 - 2a + c = 0 \quad \implies c = 2a - a^3 \quad (1)$

$b^3 - 2b + c = 0 \quad \implies c = 2b - b^3 \quad (2)$

Equating (1) and (2):

$2a - a^3 = 2b - b^3$

$b^3 - a^3 = 2b - 2a$

Factorizing both sides:

$(b-a)(b^2+ab+a^2) = 2(b-a)$

Since 'a' and 'b' are distinct, $b-a \neq 0$. We can divide both sides by $(b-a)$:

$a^2+ab+b^2 = 2 \quad (3)$

Let the given expression be $E_1 = a^2(2a^2+4ab+3b^2)$. We are given $E_1 = 3$.

Let the expression to be found be $E_2 = b^2(3a^2+4ab+2b^2)$.

Let's manipulate $E_1$ using the relation $a^2+ab+b^2=2$.

We can rewrite the term $(2a^2+4ab+3b^2)$ as follows:

$2a^2+4ab+3b^2 = 2(a^2+ab+b^2) + 2ab + b^2 = 2(2) + 2ab + b^2 = 4 + 2ab + b^2$.

So, $E_1 = a^2(4+2ab+b^2) = 4a^2 + 2a^3b + a^2b^2$.

Thus, we have $4a^2 + 2a^3b + a^2b^2 = 3 \quad (4)$.

Now, let's manipulate $E_2$ using the relation $a^2+ab+b^2=2$.

We can rewrite the term $(3a^2+4ab+2b^2)$ as follows:

$3a^2+4ab+2b^2 = 2(a^2+ab+b^2) + a^2 + 2ab = 2(2) + a^2 + 2ab = 4 + a^2 + 2ab$.

So, $E_2 = b^2(4+a^2+2ab) = 4b^2 + a^2b^2 + 2ab^3 \quad (5)$.

Notice the symmetry between equations (4) and (5).

If we swap 'a' and 'b' in equation (4), we get:

$4b^2 + 2b^3a + b^2a^2 = 3$.

This is exactly the expression for $E_2$ (equation 5).

Since the relation $a^2+ab+b^2=2$ is symmetric in 'a' and 'b', and the expression we need to find is symmetric to the given expression when 'a' and 'b' are swapped, the value of $E_2$ must be the same as $E_1$.

Therefore, $b^2(3a^2+4ab+2b^2) = 3$.