If α and β be two roots of the equation x² - 64x + 256 = 0. Then the value of (α³/β⁵)¹/⁸ + (β³/α⁵)¹/

Q: Wait, why don't we just solve for $\alpha$ and $\beta$ using the quadratic formula first?

You absolutely could! But in JEE Advanced, time is your most precious currency. Solving for roots often leads to irrational numbers that make the subsequent exponentiation a nightmare. By using Vieta's formulas, we bypass the roots entirely and work with the 'DNA' of the equation, which is much cleaner.

Q: Why did the exponents $\frac{3}{8}$ and $\frac{5}{8}$ add up to $1$ in the numerator?

That is the beauty of exponent laws! When you cross-multiply to find a common denominator, you get $\alpha^{3/8} \cdot \alpha^{5/8}$. Since the bases are the same, you add the exponents: $\frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1$. It is a perfect, elegant cancellation.

Q: Is it always safe to assume $\alpha\beta$ is positive when dealing with fractional exponents?

Great catch! Generally, you must be careful with roots of negative numbers. However, here $\alpha\beta = 256$, which is positive. If the product were negative, we would have to consider complex numbers, but for this problem, we are safely in the realm of real numbers.

Q: How did you know to rewrite $256$ as $2^8$?

Whenever you see large numbers in exponents, look for powers of $2$, $3$, or $5$. Since the exponent was $5/8$, I knew that if I could write $256$ as something to the power of $8$, the $8$s would cancel out perfectly. Always look for that 'hidden' base!

Q: What if the expression was $(\frac{\alpha^3}{\beta^5})^{1/8} - (\frac{\beta^3}{\alpha^5})^{1/8}$ instead?

The strategy remains identical! You would still find the common denominator, but the numerator would become $\alpha - \beta$. You would then use the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ to find the value. The logic holds firm!

Question

If $\alpha$ and $\beta$ be two roots of the equation $x^2 - 64x + 256 = 0$. Then the value of $(\frac{\alpha^3}{\beta^5})^{1/8} + (\frac{\beta^3}{\alpha^5})^{1/8}$ is

Options:

(A) 3

(B) 2

(C) 4

(D) 1

Sigma Percentile · Accepted Answer

### The Symphony of Roots: A Journey into Algebraic Elegance

Welcome, fellow traveler of the JEE path. Today, we are not just solving a quadratic equation; we are uncovering a hidden symmetry.

Many students look at an expression like $\left( \frac{\alpha^3}{\beta^5} ight)^{1/8} + \left( \frac{\beta^3}{\alpha^5} ight)^{1/8}$ and feel an immediate urge to panic. They see the fractional exponents, the roots of a quadratic, and the complex-looking fractions, and they freeze.

But I want you to take a deep breath. In mathematics, complexity is often just a mask for a deeper, simpler truth waiting to be revealed.

### Phase 1: The Foundation

We start with the quadratic equation $x^2 - 64x + 256 = 0$. We know that $\alpha$ and $\beta$ are its roots.

Instead of rushing to find the roots using the quadratic formula—which would lead us into a forest of square roots—we turn to the wisdom of Vieta. Vieta's formulas are the bridge between the coefficients of an equation and the properties of its roots.

We know that for any quadratic $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product is $\alpha\beta = \frac{c}{a}$.

Applying this to our equation, we find:

$$\alpha + \beta = -\frac{-64}{1} = 64$$

$$\alpha\beta = \frac{256}{1} = 256$$

These two numbers, $64$ and $256$, are the keys to our kingdom. Everything else is just algebraic manipulation.

### Phase 2: The Algebraic Dance

Now, let us look at our target expression: $\left( \frac{\alpha^3}{\beta^5} ight)^{1/8} + \left( \frac{\beta^3}{\alpha^5} ight)^{1/8}$. Using the laws of exponents, we can distribute the $1/8$ power to the numerator and denominator:

$$\frac{\alpha^{3/8}}{\beta^{5/8}} + \frac{\beta^{3/8}}{\alpha^{5/8}}$$

This looks much friendlier, doesn't it? To add these two fractions, we need a common denominator. The least common multiple of $\beta^{5/8}$ and $\alpha^{5/8}$ is simply $(\alpha\beta)^{5/8}$.

When we cross-multiply, something magical happens in the numerator:

$$	ext{Numerator} = (\alpha^{3/8} \cdot \alpha^{5/8}) + (\beta^{3/8} \cdot \beta^{5/8})$$

Recall the rule $a^m \cdot a^n = a^{m+n}$. Here, $\frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1$. The exponents vanish, leaving us with the elegant sum $\alpha + \beta$.

Our entire expression has collapsed into:

$$\frac{\alpha + \beta}{(\alpha\beta)^{5/8}}$$

### Phase 3: The Grand Finale

We have arrived at the final stage. We know $\alpha + \beta = 64$ and $\alpha\beta = 256$. Substituting these in, we get:

$$\frac{64}{(256)^{5/8}}$$

Now, look at $256$. Your mathematical intuition should immediately recognize this as $2^8$. This is no coincidence; the problem was designed with this beauty in mind.

Substituting $2^8$ for $256$ gives us:

$$\frac{64}{(2^8)^{5/8}}$$

Using the power of a power rule $(a^m)^n = a^{m \cdot n}$, the $8$ in the exponent cancels with the $8$ in the denominator of the fraction:

$$(2^8)^{5/8} = 2^{8 \cdot (5/8)} = 2^5 = 32$$

Finally, we perform the last division:

$$\frac{64}{32} = 2$$

And there it is. The **answer is 2**. We started with a daunting expression and, through the power of Vieta and the laws of exponents, we stripped away the complexity to find a simple integer.

This, my friend, is the essence of JEE Advanced mathematics: not brute force, but the elegant application of fundamental principles. Keep this clarity in your heart as you face your next challenge.

Animated Solution for Mathematics - Quadratic Equations: If $α$ and $β$ be two roots of the equation $x^{2} - 64 x + 256 = 0$ . Then the value of $(\frac{α ^{3}}{β ^{5}})^{1/8} + (\frac{β ^{3}}{α ^{5}})^{1/8}$ is

Select Answer:

Visualized Solution

Understanding the Quadratic Equation

Finding the Sum of Roots

Finding the Product of Roots

Rewriting the Target Expression

Taking the Common Denominator

Simplifying the Numerator

Finalizing the Numerator

The Simplified Expression

Substituting the Sum and Product

Rewriting 256

Simplifying the Denominator

Calculating the Final Value

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Roots

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If α and β be two roots of the equation x2−64x+256=0. Then the value of (β5α3​)1/8+(α5β3​)1/8 is