Cube of a Difference Practice Test

Use and recognize the identity (a-b)³ through expansion, factorization, equations, coefficients, and applications.

Cube of a Difference Practice Test

20 varied questions on the cube-of-a-difference identity.

Instant feedback · Worked explanations

Alternating Sign Cascade

The coefficients stay one–three–three–one. The signs alternate.

The cube of a difference is a four-term identity. The first term is positive, the first mixed term is negative, the second mixed term is positive, and the final cube is negative. At the same time, the power of the first quantity decreases while the power of the second increases.

{(u - v)}^{3} = u^{3} - 3 u^{2} v + 3 u v^{2} - v^{3}

Slot 1First quantity cubed.

Slot 2Negative three times the first square and second quantity.

Slot 3Positive three times the first quantity and second square.

Slot 4Negative second quantity cubed.

1. Track signs and powers at the same time

The total degree stays three in every term, but the signs follow a strict alternating pattern.

positive

First cube

u^{3}

The first quantity begins at power three.

negative

First mixed term

- 3 u^{2} v

The first power falls to two while the second quantity enters.

positive

Second mixed term

3 u v^{2}

The first power falls again while the second rises to power two.

negative

Final cube

- v^{3}

The second quantity finishes at power three with a negative sign.

2. Complete quantities must be substituted into all four slots

Scaled terms, variables, and fractions must keep their coefficients attached through every cube and mixed product.

Substitution matrixBuild the four terms separately before combining them.

Scaled variable

{(2 x - 3)}^{3}

8 x^{3} - 36 x^{2} + 54 x - 27

The coefficient is cubed in the first term and participates in both mixed terms.

Two variables

{(3 x - 2 y)}^{3}

27 x^{3} - 54 x^{2} y + 36 x y^{2} - 8 y^{3}

The signs alternate while the powers migrate across the four terms.

Fractional second term

{(x - \frac{1}{2})}^{3}

x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x - \frac{1}{8}

The same pattern works with rational quantities.

3. Reverse recognition reconstructs a perfect cube of a difference

Use the outer cubes to identify the binomial parts, then verify both middle coefficients and the alternating signs.

Monic perfect cube

x^{3} - 12 x^{2} + 48 x - 64 = {(x - 4)}^{3}

The outer cubes suggest the binomial; the middle terms confirm it.

Scaled perfect cube

8 x^{3} - 36 x^{2} + 54 x - 27 = {(2 x - 3)}^{3}

The leading coefficient must itself come from a perfect cube.

4. Missing coefficients are fixed by the mixed-term slots

Once the outer cubes reveal the two binomial quantities, the interior coefficients cannot vary freely.

Reconstruction rule

The magnitude of each mixed coefficient comes from coefficient three and the appropriate powers. Its sign comes from the alternating pattern.

Missing magnitude in the negative squared-variable term

x^{3} - k x^{2} + 27 x - 27

k = 3 \cdot 1 \cdot 3 = 9

Missing magnitude in the positive linear term

x^{3} - 15 x^{2} + k x - 125

k = 3 \cdot 1 \cdot 5^{2} = 75

After finding the magnitude, use the correct sign for its slot in the identity.

5. Coefficient questions can target one slot without requiring the full expansion

Identify whether the requested term comes from the first or second mixed product.

Squared-variable slot

- 3 {(4 x)}^{2} (2)

- 96 x^{2}

This is the negative second term of the four-term identity.

+ − + −⇄

Linear-variable slot

3 (4 x) {(2)}^{2}

48 x

This is the positive third term, where the second quantity is squared.

6. Cubed-binomial equations are usually simpler before expansion

Take the real cube root when the cubed form is already visible. If the polynomial is a perfect cube, recognize it first.

Direct cubed-binomial equation

{(x - 2)}^{3} = 64

x - 2 = 4

x = 6

Expansion is unnecessary because the inverse cube operation solves the structure immediately.

Polynomial recognized as a perfect cube

x^{3} - 6 x^{2} + 12 x - 8 = 27

{(x - 2)}^{3} = 27

x = 5

Reverse recognition compresses the four-term polynomial back into one cubed binomial.

7. Mental arithmetic and volume models follow the same alternating expansion

A convenient base minus a small adjustment is a natural cube-of-a-difference application.

Mental calculation just below one hundred

99^{3} = {(100 - 1)}^{3}

1000000 - 30000 + 300 - 1

= 970299

The alternating signs make the correction terms easy to organize.

Volume of a cube with a shortened edge

V = {(x - 2)}^{3}

V = x^{3} - 6 x^{2} + 12 x - 8

The four terms show exactly how the volume changes when the edge length is reduced by a fixed amount.

8. Verification can use both re-expansion and a simple numerical substitution

A correct identity must agree symbolically and numerically.

Two-pass verificationRe-expand the claimed binomial cube, then test one simple input if a quick numerical check is useful.

Symbolic check

{(x - 4)}^{3}

x^{3} - 12 x^{2} + 48 x - 64

All four terms and alternating signs return.

Numerical spot-check

{(3 - 1)}^{3} = 8

27 - 27 + 9 - 1 = 8

Both forms produce the same value for the chosen simple substitution.

9. Error analysis: the cube of a difference is neither a two-term identity nor a square pattern

Most distractors break either the four-term structure, coefficient pattern, or sign alternation.

Middle terms omitted

${(u - v)}^{3} \neq u^{3} - v^{3}$

Third term given the wrong sign

${(u - v)}^{3} \neq u^{3} - 3 u^{2} v - 3 u v^{2} - v^{3}$

One–three–three–one lost

${(u - v)}^{3} \neq u^{3} - u^{2} v + u v^{2} - v^{3}$

Coefficient not cubed

The complete monomial quantity must be cubed, including its numerical coefficient.

Variable powers reduced incorrectly

The first power decreases by one each slot while the second increases by one.

Square identity substituted by accident

A cube expansion has four terms and total degree three in every term.

Final cube-of-a-difference audit

Before accepting an expansion or factorization, check all four slots, coefficients, powers, and alternating signs.

Did I identify the two complete binomial quantities?Keep coefficients, variables, and fractions attached.

Did I use one–three–three–one?The magnitudes of the middle coefficients are both three.

Did I use the sign pattern positive, negative, positive, negative?Do not make all interior terms negative.

Did the first power decrease from three to zero?Track the first quantity across all four terms.

Did the second power increase from zero to three?The total degree remains three in every term.

For reverse recognition, did I verify all four terms?Outer cubes alone are not enough; the two mixed terms must match too.