Cube of a Sum Practice Test

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

Cube of a Sum Practice Test

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

Instant feedback · Worked explanations

Pascal Cube Loom

Four terms. One–three–three–one. Powers move in opposite directions.

The cube of a sum is easiest to control when you track two patterns at once. The coefficients follow one, three, three, one. At the same time, the power of the first term decreases from three to zero while the power of the second term increases from zero to three. Every sign stays positive.

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

Term 1First quantity cubed.

Term 2Coefficient three; first power drops by one.

Term 3Coefficient three; second power rises again.

Term 4Second quantity cubed.

1. The power pattern is as important as the coefficient pattern

Across the four terms, one exponent decreases while the other increases, and the total degree remains three.

First cube

u^{3}

The first quantity begins at power three.

First mixed term

3 u^{2} v

The first power drops to two and the second enters at power one.

Second mixed term

3 u v^{2}

The first power drops to one while the second rises to two.

Second cube

v^{3}

The second quantity finishes at power three.

2. Substitute complete quantities into the four slots

Coefficients, variables, and rational terms must remain attached to the quantity being cubed.

Substitution deckBuild all four terms separately before combining anything.

Scaled variable

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

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

The coefficient on the variable is cubed and also appears in both middle terms.

Two variables

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

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

Both variables follow the same descending-and-ascending power pattern.

Rational second term

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

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

Fractions are treated as complete algebraic quantities inside the identity.

3. Reverse recognition identifies a perfect binomial cube from four terms

First cube-root the outer terms. Then verify that the two middle terms match the one–three–three–one pattern.

Monic perfect cube

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

The outer cubes identify the two binomial parts; the middle coefficients confirm the match.

Scaled perfect cube

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

The leading coefficient must also be a perfect cube to rebuild the scaled first binomial part.

4. Missing coefficients are determined by the two middle slots

Once the outer cubes reveal the two binomial parts, the mixed-term coefficients are no longer arbitrary.

Coefficient reconstruction

Use the outer cubes to recover the two binomial quantities, then calculate the appropriate mixed term with coefficient three.

Missing coefficient on the squared-variable term

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

k = 3 \cdot 1 \cdot 3 = 9

Missing coefficient on the linear term

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

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

The first mixed term squares the first quantity; the second mixed term squares the second.

5. If a question asks for one coefficient, calculate only the relevant slot

You do not always need the complete expansion.

Coefficient of the squared-variable term

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

96 x^{2}

This comes from the second slot of the cube identity.

1–3–3–1⇄

Coefficient of the linear-variable term

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

48 x

This comes from the third slot, where the second quantity is squared.

6. Cubed-binomial equations can often be solved before any expansion

If the cube structure is already visible, take the real cube root first. If a four-term polynomial is a perfect cube, recognize it before solving.

Already in cubed-binomial form

{(x + 2)}^{3} = 125

x + 2 = 5

x = 3

Expanding would create unnecessary work.

Polynomial recognized as a cube

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

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

x = 1

Reverse recognition converts the equation back to a simple cubed binomial.

7. Numerical and geometric applications use exactly the same four-term structure

The identity is useful whenever a quantity is naturally written as a convenient base plus an adjustment.

Mental calculation near one hundred

101^{3} = {(100 + 1)}^{3}

1000000 + 30000 + 300 + 1

= 1030301

The four terms are much easier to evaluate than long multiplication of the original cube.

Volume of an enlarged cube

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

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

The four terms describe how the volume changes when each edge length is increased by the same amount.

8. Re-expansion verifies a proposed perfect-cube factorization

Check all four terms, both middle coefficients, and the power migration.

Reverse-check stripA correct repeated binomial cube must reproduce the exact four-term polynomial.

{(x + 4)}^{3}

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

The coefficients, powers, and all positive signs match the cube-of-a-sum identity.

9. Error analysis: most wrong expansions break either the middle terms or the power pattern

A cube of a sum has four terms, not two or three.

Middle terms omitted

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

Square identity used instead

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

Coefficient not cubed

The complete first or second quantity must be cubed, including its numerical coefficient.

Factor three lost

Both mixed terms carry coefficient three.

Signs changed unnecessarily

Every term in a cube-of-a-sum expansion is positive.

Mixed powers swapped

The second term squares the first quantity; the third term squares the second.

Final cube-of-a-sum audit

Before accepting the expansion or factorization, check all four coefficient and power slots.

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

Did I use the coefficient pattern one–three–three–one?The two middle terms both carry coefficient three.

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 of each term remains three.

Did I keep every sign positive?This page is specifically the cube of a sum.

For reverse factoring, did I re-expand to verify all four terms?Outer cubes alone are not enough; both middle terms must match too.