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Answer
The expression \( 27u^{3} + w^{3} \) factors to \( (3u + w)(9u^{2} - 3uw + w^{2}) \).
Solution
1. Recognize that \( 27u^3 \) and \( w^3 \) are perfect cubes because:
\[
27u^3 = (3u)^3 \quad \text{and} \quad w^3 = w^3.
\]
2. Write the expression as a sum of cubes:
\[
27u^3 + w^3 = (3u)^3 + w^3.
\]
3. Use the sum of cubes formula:
\[
a^3+b^3 = (a+b)(a^2-ab+b^2).
\]
Here, \( a = 3u \) and \( b = w \).
4. Substitute into the formula:
\[
(3u)^3+w^3 = (3u+w)\left((3u)^2-3u\cdot w+w^2\right).
\]
5. Simplify the expression inside the parentheses:
\[
(3u)^2 = 9u^2,
\]
so the factorization becomes:
\[
(3u+w)(9u^2-3uw+w^2).
\]
Thus, the factorization of \( 27u^3+w^3 \) is:
\[
(3u+w)(9u^2-3uw+w^2).
\]
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Simplify this solution The Deep Dive
To factor \( 27u^3 + w^3 \), we can recognize this as a sum of cubes. It can be expressed as: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \] where \( a = 3u \) and \( b = w \). So, \[ 27u^3 + w^3 = (3u)^3 + w^3 \] Applying the formula, we get: \[ (3u + w)((3u)^2 - (3u)w + w^2) \] This simplifies to: \[ (3u + w)(9u^2 - 3uw + w^2) \] Putting it all together, the factored form is: \[ (3u + w)(9u^2 - 3uw + w^2) \] Now you've got a neat factored expression!
