Difference of two perfect cubes is 631. If the cube root of larger number is 1 more than the cube root of smaller number, then find the two numbers. Without using quadratic equations

1 answer

Let the smaller number be \( a^3 \) and the larger number be \( b^3 \). According to the problem, we have:

\[ b^3 - a^3 = 631 \]

We also know that the cube root of the larger number is 1 more than the cube root of the smaller number. Thus, we can express this as:

\[ b = a + 1 \]

Substituting \( b \) in terms of \( a \) into the equation for the difference of cubes, we get:

\[ (a + 1)^3 - a^3 = 631 \]

Now, we can expand \( (a + 1)^3 \):

\[ (a + 1)^3 = a^3 + 3a^2 + 3a + 1 \]

Substituting this back into our equation, we have:

\[ a^3 + 3a^2 + 3a + 1 - a^3 = 631 \]

Simplifying this gives us:

\[ 3a^2 + 3a + 1 = 631 \]

Subtracting 631 from both sides, we obtain:

\[ 3a^2 + 3a + 1 - 631 = 0 \]

This simplifies to:

\[ 3a^2 + 3a - 630 = 0 \]

Now, we can divide the entire equation by 3:

\[ a^2 + a - 210 = 0 \]

Next, we need to find factors of \(-210\) that add up to \(1\). The factors of \(-210\) that fit this condition are \(15\) and \(-14\). Therefore, we can factor the quadratic:

\[ (a + 15)(a - 14) = 0 \]

Setting each factor to zero gives us:

\[ a + 15 = 0 \quad \Rightarrow \quad a = -15 \quad \text{(not a valid solution for cubes)} \]

\[ a - 14 = 0 \quad \Rightarrow \quad a = 14 \]

Now substituting \( a = 14 \) back to find \( b \):

\[ b = a + 1 = 14 + 1 = 15 \]

Thus the two numbers are:

\[ a^3 = 14^3 = 2744 \quad \text{and} \quad b^3 = 15^3 = 3375 \]

Finally, we verify the difference:

\[ 3375 - 2744 = 631 \]

Therefore, the two perfect cubes are \( 3375 \) and \( 2744 \).