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Sum of cubes (factored)
Toolkit 12
Sum of cubes (factored)
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
a^3+b^3=(a+b)(a^2-ab+b^2)
a
3
+
b
3
=
(
a
+
b
)
(
a
2
−
ab
+
b
2
)
Proof
Using the distributive law,
(
a
+
b
)
(
a
2
−
a
b
+
b
2
)
=
a
3
−
a
2
b
+
a
b
2
+
a
2
b
−
a
b
2
+
b
3
=
a
3
+
b
3
.
□
\begin{aligned} (a+b)(a^2 - ab + b^2) &= a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 \\ &= a^3 + b^3. \quad\square \end{aligned}
(
a
+
b
)
(
a
2
−
ab
+
b
2
)
=
a
3
−
a
2
b
+
a
b
2
+
a
2
b
−
a
b
2
+
b
3
=
a
3
+
b
3
.
□
Related Problems
Core
AMC 10A 2020 (Problem 14)
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