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If one real root of the quadratic equation 81x^2 kx 256 =0 is cube of the other root, then a value of k is?
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If one real root of the quadratic equation 81x^2 kx 256 =0 is cube of ...
**Solution:**

Let's solve the given quadratic equation step by step to find the value of k.

The given quadratic equation is: 81x^2 + kx + 256 = 0

**Step 1: Finding the roots of the quadratic equation**

We can find the roots of a quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Here, a = 81, b = k, and c = 256

Substituting these values into the quadratic formula, we get:

x = (-k ± √(k^2 - 4*81*256)) / 2*81

Simplifying further:

x = (-k ± √(k^2 - 82944)) / 162

**Step 2: Given condition of the roots**

According to the given condition, one root of the quadratic equation is the cube of the other root. Let's assume the roots as p and q, where q = p^3.

So, we have two equations:

p = q^3 ----(1)

p + q = -k/81 ----(2)

**Step 3: Solving the equations**

Substituting q = p^3 from equation (1) into equation (2), we get:

p + p^3 = -k/81

p^3 + p + k/81 = 0

Multiplying the equation by 81 to eliminate the fraction, we have:

81p^3 + 81p + k = 0 ----(3)

**Step 4: Comparing the coefficients**

Now, let's compare the coefficients of the given quadratic equation (81x^2 + kx + 256 = 0) and equation (3) (81p^3 + 81p + k = 0) to find the value of k.

Comparing the coefficients of p^3:

81 = 0

Since the coefficients do not match, we can conclude that the given quadratic equation does not satisfy the given condition.

Therefore, there is no specific value of k that satisfies the given condition.

Hence, there is no solution for the value of k.
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If one real root of the quadratic equation 81x^2 kx 256 =0 is cube of the other root, then a value of k is?
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