Understanding the Problem: Solving for a Cubic Expression

We are given two equations involving variables \(x\) and \(y\):

• Equation 1: \(5x + y = 17\)
• Equation 2: \(xy = 6\)

Our goal is to find the value of the expression \(125x^3 + y^3\).

Connecting the Expression to Known Algebraic Identities

The expression \(125x^3 + y^3\) can be rewritten as \((5x)^3 + y^3\). This form looks like the sum of two cubes. Recall the algebraic identity for the sum of cubes:

\(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)

In our case, if we let \(a = 5x\) and \(b = y\), the identity becomes:

\((5x)^3 + y^3 = (5x + y)((5x)^2 - (5x)(y) + y^2)\)

Simplifying the terms inside the second parenthesis:

\(125x^3 + y^3 = (5x + y)(25x^2 - 5xy + y^2)\)

Using the Given Equations to Find Necessary Values

From the given equations, we already know the values for some parts of the expanded expression:

• We know \(5x + y = 17\) from Equation 1.
• We know \(xy = 6\) from Equation 2, which means \(5xy = 5 \times 6 = 30\).

However, we still need the value of \(25x^2 + y^2\).

Finding the Value of \(25x^2 + y^2\)

We can find \(25x^2 + y^2\) by using the square of the first equation. Recall the identity for squaring a binomial:

\((a+b)^2 = a^2 + 2ab + b^2\)

Applying this to our first equation, \((5x + y) = 17\):

\((5x + y)^2 = (17)^2\)

\((5x)^2 + 2(5x)(y) + y^2 = 289\)

\(25x^2 + 10xy + y^2 = 289\)

Now, we can rearrange this equation to isolate \(25x^2 + y^2\):

\(25x^2 + y^2 = 289 - 10xy\)

Substitute the value of \(xy = 6\) into this equation:

\(25x^2 + y^2 = 289 - 10(6)\)

\(25x^2 + y^2 = 289 - 60\)

\(25x^2 + y^2 = 229\)

Calculating the Final Value of the Expression

Now we have all the components needed to find the value of \(125x^3 + y^3\) using the expanded form derived earlier:

\(125x^3 + y^3 = (5x + y)(25x^2 - 5xy + y^2)\)

Substitute the values we found:

• \(5x + y = 17\)
• \(25x^2 + y^2 = 229\)
• \(5xy = 30\)

Plugging these values into the expression:

\(125x^3 + y^3 = (17)((25x^2 + y^2) - 5xy)\)

\(125x^3 + y^3 = (17)(229 - 30)\)

\(125x^3 + y^3 = (17)(199)\)

Finally, perform the multiplication:

\(17 \times 199 = 17 \times (200 - 1)\)

\(17 \times 199 = 17 \times 200 - 17 \times 1\)

\(17 \times 199 = 3400 - 17\)

\(17 \times 199 = 3383\)

So, the value of \(125x^3 + y^3\) is 3383.

Summary of Steps

Revision Table: Key Algebraic Identities

Additional Information: Solving Systems of Equations

While we used algebraic identities to solve this problem, it is also possible to find the individual values of \(x\) and \(y\) by solving the system of equations:

• \(5x + y = 17\)
• \(xy = 6\)

From the first equation, we can express \(y\) in terms of \(x\): \(y = 17 - 5x\).

Substitute this into the second equation:

\(x(17 - 5x) = 6\)

\(17x - 5x^2 = 6\)

\(5x^2 - 17x + 6 = 0\)

This is a quadratic equation that can be solved for \(x\) using factoring or the quadratic formula. Let's try factoring:

We need two numbers that multiply to \(5 \times 6 = 30\) and add up to -17. These numbers are -15 and -2.

\(5x^2 - 15x - 2x + 6 = 0\)

\(5x(x - 3) - 2(x - 3) = 0\)

\((5x - 2)(x - 3) = 0\)

This gives two possible values for \(x\):

• \(x - 3 = 0 \implies x = 3\)
• \(5x - 2 = 0 \implies 5x = 2 \implies x = \frac{2}{5}\)

Now find the corresponding values of \(y\) using \(y = 17 - 5x\):

• If \(x = 3\), then \(y = 17 - 5(3) = 17 - 15 = 2\). Let's check \(xy = 3 \times 2 = 6\). This solution works.
• If \(x = \frac{2}{5}\), then \(y = 17 - 5\left(\frac{2}{5}\right) = 17 - 2 = 15\). Let's check \(xy = \frac{2}{5} \times 15 = 2 \times 3 = 6\). This solution also works.

We can use either pair of \((x, y)\) values to find \(125x^3 + y^3\).

Case 1: \(x=3, y=2\)

\(125x^3 + y^3 = 125(3)^3 + (2)^3 = 125(27) + 8\)

\(125 \times 27 = 125 \times (30 - 3) = 3750 - 375 = 3375\)

\(125x^3 + y^3 = 3375 + 8 = 3383\)

Case 2: \(x=\frac{2}{5}, y=15\)

\(125x^3 + y^3 = 125\left(\frac{2}{5}\right)^3 + (15)^3 = 125\left(\frac{8}{125}\right) + 3375\)

\(125x^3 + y^3 = 8 + 3375 = 3383\)

Both pairs give the same result, confirming our previous calculation using algebraic identities.