Find one pair of real numbers, (x,y), such that x + y = 6 and x^3 + y^3 = 144.

To find a pair of real numbers (x, y) that satisfy the given equations, we can use a system of equations method.

We are given two equations:
1) x + y = 6
2) x^3 + y^3 = 144

Step 1: Solve equation 1 to express one variable in terms of the other.
From equation 1, we have x = 6 - y.

Step 2: Substitute the value of x from Step 1 into equation 2.
Replacing x in equation 2, we get (6 - y)^3 + y^3 = 144.

Step 3: Expand and simplify the equation obtained in Step 2.
Expanding (6 - y)^3, we have (6 - y)(6 - y)(6 - y) = 144.
Simplifying, we get (216 - 36y + y^2)(6 - y) = 144.
Multiplying, we have 1296 - 216y + 36y^2 - 36y + 6y^2 - y^3 = 144.

Step 4: Rearrange the equation to its simplest form.
Combining like terms, we get -y^3 + 42y^2 - 252y + 1152 = 0.

Step 5: Solve the equation obtained in Step 4.
To solve this equation, we can use numerical methods like factoring, synthetic division, or graphing calculators. After solving, we find that one possible real solution is y = 4.

Step 6: Substitute the value of y into the expression obtained in Step 1 to solve for x.
Substituting y = 4 into x = 6 - y, we get x = 6 - 4 = 2.

Therefore, one pair of real numbers that satisfy the given equations is (x, y) = (2, 4).

x^3+y^3 = (x+y)(x^2-xy+y^2)

6(x^2-xy+y^2) = 144
x^2-xy+y^2 = 24

y = 6-x, so

x^2-x(6-x)+(6-x)^2 = 24
x^2-6x+x^2+x^2-12x+36 = 24
3x^2-18x+12 = 0
3(x^2-6x+4) = 0
Look familiar?