Factor completely.y3−127y^3-\frac{1}{27}

( y − 1 3 ) ( y 2 + 1 3 y + 1 9 ) \left(y-\frac13\right)\left(y^2+\frac13y+\frac19\right)

Factor completely. y 3 − 1 27 y^3-\frac{1}{27}

Here we want to factor the polynomial y cubed minus one over 27. And since we have a variable being cubed we suspect this might be a difference of cubes. So let's check. We know y cubed is a cube term as it has the factors y times y times y. And we know 27 is a cube term as it has the factors three times three times three. And one is also a cubed term as it has the factors one times one times one. So this is a difference of cubes. And we can rewrite this as y cubed minus one third cubed. Now to factor we'll have a, which is y, and then our same sign, so minus b, which is one third, times a squared, so y squared, and then opposite, so positive, plus AB, so y times one third or one third y. And then our final sine value is always positive, so plus b squared, one third squared. Simplifying, we get y minus one third times y squared plus one third y plus one ninth, and this is our final answer.

Factor completely. y 3 − 1 27 y^3-\frac{1}{27}

( y − 1 3 ) ( y 2 + 1 3 y + 1 9 ) \left(y-\frac13\right)\left(y^2+\frac13y+\frac19\right)

( y + 1 3 ) ( y 2 + 1 3 y + 1 9 ) \left(y+\frac13\right)\left(y^2+\frac13y+\frac19\right)

y ( y − 1 3 ) 2 y\left(y-\frac13\right)^2 y ( y − 3 1 ) 2

− y ( y + 1 3 ) 2 -y\left(y+\frac13\right)^2 − y ( y + 3 1 ) 2

7. Factoring

Factoring Special Products

Intermediate Algebra

7. Factoring

Factoring Special Products

Struggling with Intermediate Algebra?

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Multiple Choice

Factor completely.
$y^{3} - \frac{1}{27}$

$(y + \frac{1}{3}) (y^{2} + \frac{1}{3} y + \frac{1}{9})$

$(y - \frac{1}{3}) (y^{2} + \frac{1}{3} y + \frac{1}{9})$

$y$\left$(y-$\frac$13$\right$)^2$

$-y$\left$(y+$\frac$13$\right$)^2$

Verified step by step guidance

Identify whether the given expression is a sum or difference of cubes. Recall that a sum of cubes has the form $a^3 + b^3$ and a difference of cubes has the form $a^3 - b^3$.

Express each term in the expression as a cube of a simpler term, i.e., find $a$ and $b$ such that the expression matches either $a^3 + b^3$ or $a^3 - b^3$.

Use the appropriate factoring formula: for sum of cubes, use $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$; for difference of cubes, use $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.

Substitute the values of $a$ and $b$ into the formula and write the factored form as the product of a binomial and a trinomial.

Check your factorization by expanding the factors to ensure they multiply back to the original expression.

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Struggling with Intermediate Algebra?

Factor completely.y3−127y^3-\(\frac{1}{27}\)

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Factor completely.
$y^{3} - \frac{1}{27}$