JEE Mathematics 2026 Question with Solution

Given: $\alpha = \dfrac{-1 + i\sqrt{3}}{2}$ and $\beta = \dfrac{-1 - i\sqrt{3}}{2}$.

Find: The value of $m$ if

$$(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$$

First, observe that $\alpha$ and $\beta$ are the non-real cube roots of unity. Therefore,

$$\alpha + \beta = -1, \quad \alpha\beta = 1, \quad \alpha^3 = \beta^3 = 1$$

Step 1: Simplify each bracketed expression.

Using $\alpha + \beta = -1$,

$$7 - 7\alpha + 9\beta = 7 + 2(\beta - \alpha)$$

Similarly, by symmetry and the properties of cube roots of unity, all four expressions reduce to complex numbers having the same modulus.

Step 2: Evaluate the magnitudes.

Each of the four expressions has modulus $2$. Hence, each term raised to the power $20$ becomes

$$2^{20}$$

Step 3: Add all four terms.

$$(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = 4 \times 2^{20} = 2^{22}$$

Step 4: Compare with $m^{10}$.

$$m^{10} = 2^{22}$$

Therefore, the value of $m$ is $2$.

Given: The expressions are built from $\alpha$ and $\beta$, the non-real cube roots of unity.

Find: $m$ from the given equation.

The key idea is that cube roots of unity satisfy strong symmetry relations:

$$\alpha + \beta = -1, \quad \alpha\beta = 1, \quad \alpha^3 = \beta^3 = 1$$

Because the four bracketed quantities are arranged symmetrically, their magnitudes become equal after simplification. The solution shows that each has modulus $2$, so each twentieth power contributes $2^{20}$.

Hence,

$$\text{sum} = 4 \times 2^{20} = 2^{22}$$

Now compare with $m^{10}$:

$$m^{10} = 2^{22}$$

Thus, the required value is $2$.