Factor d^(3)-1 completely. Answer:

Recognize Cubes Difference: Recognize that d^3 - 1 is a difference of cubes. A difference of cubes can be factored using the formula a^3 - b^3 = (a - b)(a^2 + ab + b^2). Here, a = d and b = 1.Apply Formula: Apply the difference of cubes formula. Using the formula from Step 1, we get: d^3 - 1^3 = (d - 1)(d^2 + d*1 + 1^2).Simplify Expression: Simplify the expression. Simplify the terms inside the parentheses: (d - 1)(d^2 + d + 1).Check Further Factorization: Check for any further factorization. The quadratic d^2 + d + 1 cannot be factored further over the real numbers, so the expression is fully factored.

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Factor $d^{3}-1$ completely. $\newline$ Answer:

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Math Problems

Grade 7

Powers with negative bases

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Q. Factor

d^{3}-1

completely.

\newline

Answer:

Recognize Cubes Difference: Recognize that $d^3 - 1$ is a difference of cubes. A difference of cubes can be factored using the formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ . Here, $a = d$ and $b = 1$ .
Apply Formula: Apply the difference of cubes formula. $\newline$ Using the formula from Step $1$ , we get: $\newline$ $d^3 - 1^3 = (d - 1)(d^2 + d\cdot 1 + 1^2)$ .
Simplify Expression: Simplify the expression. $\newline$ Simplify the terms inside the parentheses: $\newline$ $(d - 1)(d^2 + d + 1)$ .
Check Further Factorization: Check for any further factorization. $\newline$ The quadratic $d^2 + d + 1$ cannot be factored further over the real numbers, so the expression is fully factored.