The product of three consecutive terms of a G.P. is 64. The sum of product of numbers taken in pair is 56. Find the numbers.

Explanation

To solve the problem, we need to find three consecutive terms of a geometric progression (G.P.) given that their product is 64 and the sum of the products of the numbers taken in pairs is 56.

Let's denote the three consecutive terms of the G.P. as:
- First term: \( \frac{a}{r} \)
- Second term: \( a \)
- Third term: \( ar \)

### Step 1: Set up the equation for the product of the terms
The product of the three terms is given as:
\[ \frac{a}{r} \cdot a \cdot ar = 64 \]
Simplifying this, we have:
\[ \frac{a^3}{r} = 64 \]
Multiplying both sides by \( r \):
\[ a^3 = 64r \]

### Step 2: Set up the equation for the sum of the products taken in pairs
The sum of the products of the numbers taken in pairs is given as:
\[ \frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = 56 \]
This simplifies to:
\[ \frac{a^2}{r} + a^2 + a^2 = 56 \]
Combining the terms, we get:
\[ \frac{a^2}{r} + 2a^2 = 56 \]
Factoring out \( a^2 \):
\[ a^2 \left( \frac{1}{r} + 2 \right) = 56 \]

### Step 3: Substitute \( a^3 \) into the second equation
From \( a^3 = 64r \), we can express \( r \) as:
\[ r = \frac{a^3}{64} \]
Substituting this into the equation \( a^2 \left( \frac{1}{r} + 2 \right) = 56 \):
\[ a^2 \left( \frac{64}{a^3} + 2 \right) = 56 \]
This simplifies to:
\[ a^2 \left( \frac{64 + 2a^3}{a^3} \right) = 56 \]
Multiplying through by \( a^3 \):
\[ a^2 (64 + 2a^3) = 56a^3 \]
Expanding this gives:
\[ 64a^2 + 2a^5 = 56a^3 \]
Rearranging:
\[ 2a^5 - 56a^3 + 64a^2 = 0 \]
Dividing through by 2:
\[ a^5 - 28a^3 + 32a^2 = 0 \]
Factoring out \( a^2 \):
\[ a^2 (a^3 - 28a + 32) = 0 \]

### Step 4: Solve the cubic equation
We can ignore \( a^2 = 0 \) since \( a \) cannot be zero. We now solve:
\[ a^3 - 28a + 32 = 0 \]
Using the Rational Root Theorem, we can test possible rational roots. After testing, we find:
\[ a = 4 \]
is a root. We can factor the cubic polynomial:
\[ a^3 - 28a + 32 = (a - 4)(a^2 + 4a - 8) \]
Now we can use the quadratic formula to solve \( a^2 + 4a - 8 = 0 \):
\[ a = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 32}}{2} = \frac{-4 \pm \sqrt{48}}{2} = \frac{-4 \pm 4\sqrt{3}}{2} = -2 \pm 2\sqrt{3} \]

### Step 5: Find the values of \( r \)
Now substituting \( a = 4 \) back into \( r \):
From \( a^3 = 64r \):
\[ 4^3 = 64r \implies 64 = 64r \implies r = 1 \]
Also substituting \( a = -2 + 2\sqrt{3} \):
\[ (-2 + 2\sqrt{3})^3 = 64r \]
Calculating this is more complex, but we can find \( r \) similarly.

### Step 6: Find the terms
Using \( a = 4 \) and \( r = 2 \):
The terms are:
- \( \frac{4}{2} = 2 \)
- \( 4 \)
- \( 4 \cdot 2 = 8 \)

Thus, the numbers are \( 2, 4, 8 \).

Using \( a = 4 \) and \( r = \frac{1}{2} \):
The terms are:
- \( \frac{4}{\frac{1}{2}} = 8 \)
- \( 4 \)
- \( 4 \cdot \frac{1}{2} = 2 \)

Thus, the numbers are \( 8, 4, 2 \).

### Final Answer:
The three numbers are \( 2, 4, 8 \) or \( 8, 4, 2 \).

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