Answer :
To solve for [tex]\( x^3 - y^3 \)[/tex] given the equations [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex], let's perform the following steps:
1. Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = y + 4 \][/tex]
2. Substitute [tex]\( x = y + 4 \)[/tex] into the product equation [tex]\( xy = 21 \)[/tex]:
[tex]\[ (y + 4)y = 21 \][/tex]
[tex]\[ y^2 + 4y = 21 \][/tex]
[tex]\[ y^2 + 4y - 21 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( y^2 + 4y - 21 = 0 \)[/tex]:
The solutions can be found using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex].
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-4 + 10}{2} = 3 \quad \text{or} \quad y = \frac{-4 - 10}{2} = -7 \][/tex]
4. Find corresponding values for [tex]\( x \)[/tex]:
[tex]\[ \text{If } y = 3, \text{ then } x = y + 4 = 3 + 4 = 7 \][/tex]
[tex]\[ \text{If } y = -7, \text{ then } x = y + 4 = -7 + 4 = -3 \][/tex]
This gives us two pairs of solutions [tex]\((x, y)\)[/tex]:
[tex]\[ (x, y) = (7, 3) \quad \text{or} \quad (x, y) = (-3, -7) \][/tex]
5. Determine [tex]\( x^3 - y^3 \)[/tex] for each pair:
Recall the identity for the difference of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
Substitute [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex] into the identity. Let's calculate [tex]\( x^2 + xy + y^2 \)[/tex] for the pairs [tex]\((7, 3)\)[/tex] and [tex]\((-3, -7)\)[/tex]:
- For [tex]\((x, y) = (7, 3)\)[/tex]:
[tex]\[ x^2 = 49, \quad y^2 = 9, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 49 + 21 + 9 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
- For [tex]\((x, y) = (-3, -7)\)[/tex]:
[tex]\[ x^2 = 9, \quad y^2 = 49, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 9 + 21 + 49 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
Hence, for both pairs, [tex]\( x^3 - y^3 \)[/tex] equals 316.
Therefore, the solution is:
[tex]\[ \boxed{316} \][/tex]
1. Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ x = y + 4 \][/tex]
2. Substitute [tex]\( x = y + 4 \)[/tex] into the product equation [tex]\( xy = 21 \)[/tex]:
[tex]\[ (y + 4)y = 21 \][/tex]
[tex]\[ y^2 + 4y = 21 \][/tex]
[tex]\[ y^2 + 4y - 21 = 0 \][/tex]
3. Solve the quadratic equation [tex]\( y^2 + 4y - 21 = 0 \)[/tex]:
The solutions can be found using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -21 \)[/tex].
[tex]\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-21)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 + 84}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{100}}{2} \][/tex]
[tex]\[ y = \frac{-4 \pm 10}{2} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-4 + 10}{2} = 3 \quad \text{or} \quad y = \frac{-4 - 10}{2} = -7 \][/tex]
4. Find corresponding values for [tex]\( x \)[/tex]:
[tex]\[ \text{If } y = 3, \text{ then } x = y + 4 = 3 + 4 = 7 \][/tex]
[tex]\[ \text{If } y = -7, \text{ then } x = y + 4 = -7 + 4 = -3 \][/tex]
This gives us two pairs of solutions [tex]\((x, y)\)[/tex]:
[tex]\[ (x, y) = (7, 3) \quad \text{or} \quad (x, y) = (-3, -7) \][/tex]
5. Determine [tex]\( x^3 - y^3 \)[/tex] for each pair:
Recall the identity for the difference of cubes:
[tex]\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \][/tex]
Substitute [tex]\( x - y = 4 \)[/tex] and [tex]\( xy = 21 \)[/tex] into the identity. Let's calculate [tex]\( x^2 + xy + y^2 \)[/tex] for the pairs [tex]\((7, 3)\)[/tex] and [tex]\((-3, -7)\)[/tex]:
- For [tex]\((x, y) = (7, 3)\)[/tex]:
[tex]\[ x^2 = 49, \quad y^2 = 9, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 49 + 21 + 9 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
- For [tex]\((x, y) = (-3, -7)\)[/tex]:
[tex]\[ x^2 = 9, \quad y^2 = 49, \quad xy = 21 \][/tex]
[tex]\[ x^2 + xy + y^2 = 9 + 21 + 49 = 79 \][/tex]
[tex]\[ x^3 - y^3 = 4 \cdot 79 = 316 \][/tex]
Hence, for both pairs, [tex]\( x^3 - y^3 \)[/tex] equals 316.
Therefore, the solution is:
[tex]\[ \boxed{316} \][/tex]