Polynomial Identities (Advanced & Applications)

1) Expand (2x + 3y)³.

8x³ + 12x²y + 6xy² + 27y³

8x³ + 36x²y + 54xy² + 27y³

8x³ + 18x²y + 27y³

8x³ + 24x²y + 36xy² + 27y³

Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ makes expansion straightforward.

Substituting a = 2x and b = 3y shows how each term becomes a combination of powers of x and y.

This produces 8x³ + 36x²y + 54xy² + 27y³ by following the binomial pattern.

Explanation

Using the identity (a + b)³ = a³ + 3a²b + 3ab² + b³ makes expansion straightforward.

Substituting a = 2x and b = 3y shows how each term becomes a combination of powers of x and y.

This produces 8x³ + 36x²y + 54xy² + 27y³ by following the binomial pattern.

2) Expand (x − 2y)³.

X³ − 6x²y + 12xy² − 8y³

X³ − 8y³

X³ + 6x²y − 12xy² + 8y³

X³ − 3x²y + 3xy² − y³

The formula (a − b)³ = a³ − 3a²b + 3ab² − b³ applies here with a = x and b = 2y. Each substituted term preserves the alternating signs required by the identity. This creates x³ − 6x²y + 12xy² − 8y³.

Explanation

The formula (a − b)³ = a³ − 3a²b + 3ab² − b³ applies here with a = x and b = 2y. Each substituted term preserves the alternating signs required by the identity. This creates x³ − 6x²y + 12xy² − 8y³.

3) Simplify (x² + 3)².

X⁴ + 9

X⁴ + 3x² + 6

X⁴ + 6x² + 9

X⁴ − 6x² + 9

Use the square identity (a + b)² = a² + 2ab + b². Here, a = x² and b = 3, so after squaring each term and multiplying the middle product, you obtain x⁴ + 6x² + 9. This method ensures all cross-terms appear correctly.

Explanation

Use the square identity (a + b)² = a² + 2ab + b². Here, a = x² and b = 3, so after squaring each term and multiplying the middle product, you obtain x⁴ + 6x² + 9. This method ensures all cross-terms appear correctly.

4) Evaluate (x + 2)³ − (x − 2)³.

12x² + 16

8x² + 12

16x² + 12

Use a³ − b³ = (a − b)(a² + ab + b²) with a = x + 2 and b = x − 2.

Compute a − b = 4, then evaluate a² + ab + b², which simplifies to 3x² + 4.

Multiplying gives 12x² + 16, showing how symmetric cube terms cancel neatly.

Explanation

Use a³ − b³ = (a − b)(a² + ab + b²) with a = x + 2 and b = x − 2.

Compute a − b = 4, then evaluate a² + ab + b², which simplifies to 3x² + 4.

Multiplying gives 12x² + 16, showing how symmetric cube terms cancel neatly.

5) Verify the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

False

True

Sometimes true

True only if a + b + c = 0

Expanding (a + b + c)(a + b + c) shows every pair ab, bc, ca appears twice.

This produces the required 2(ab + bc + ca), confirming the identity.

The formula holds for all real values of a, b, and c.

Explanation

Expanding (a + b + c)(a + b + c) shows every pair ab, bc, ca appears twice.

This produces the required 2(ab + bc + ca), confirming the identity.

The formula holds for all real values of a, b, and c.

6) Expand (x + y + z)².

X² + y² + z² + 2(xy + yz + zx)

X² + y² + z² − 2(xy + yz + zx)

X² + y² + z²

2(x² + y² + z²)

When squaring a trinomial, each variable multiplies with every other, producing both square terms and pairwise products. This happens because xy, yz, and zx all appear in two multiplication paths. Thus, the full expansion is x² + y² + z² + 2(xy + yz + zx). This expansion reflects the structure of multiplying identical trinomials.

Explanation

When squaring a trinomial, each variable multiplies with every other, producing both square terms and pairwise products. This happens because xy, yz, and zx all appear in two multiplication paths. Thus, the full expansion is x² + y² + z² + 2(xy + yz + zx). This expansion reflects the structure of multiplying identical trinomials.

7) Simplify (x² − y²)².

X⁴ − 2x²y² + y⁴

X⁴ + y⁴

X⁴ + 2x²y² + y⁴

X² − y²

Treat the expression as a binomial square (a − b)² with a = x² and b = y².

Apply a² − 2ab + b² to obtain x⁴ − 2x²y² + y⁴.

The structure reflects a symmetric fourth-degree polynomial. This yields a fourth-degree polynomial showing symmetry between xxx and yyy.

Explanation

Treat the expression as a binomial square (a − b)² with a = x² and b = y².

Apply a² − 2ab + b² to obtain x⁴ − 2x²y² + y⁴.

The structure reflects a symmetric fourth-degree polynomial. This yields a fourth-degree polynomial showing symmetry between xxx and yyy.

8) If a³ + b³ = 343 and a + b = 7, find ab.

0

12

14

16343 = 7³ − 3ab(7) ⇒ 343 = 343 − 21ab ⇒ ab = 14.

Use the identity: a³ + b³ = (a + b)³ − 3ab(a + b).

Substituting values gives 343 = 343 − 21ab, so ab = 14.

This demonstrates how cube sums relate directly to the product ab. Substituting the known values creates a simple equation involving ababab. Solving for ababab shows how cube sums relate directly to the product of variables.

Explanation

Use the identity: a³ + b³ = (a + b)³ − 3ab(a + b).

Substituting values gives 343 = 343 − 21ab, so ab = 14.

This demonstrates how cube sums relate directly to the product ab. Substituting the known values creates a simple equation involving ababab. Solving for ababab shows how cube sums relate directly to the product of variables.

9) (a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a).

True

False

The proposed formula is incorrect. The true identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Comparing expansions shows the given expression does not match the real cubic identity.

Explanation

The proposed formula is incorrect. The true identity is a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca). Comparing expansions shows the given expression does not match the real cubic identity.

10) (x + y)³ + (x − y)³ = 2x³ + 6xy².

True

False

Expanding each cube reveals that +3x²y and −3x²y cancel. The terms +3xy² from both expansions add together, doubling their coefficient. This creates 2x³ + 6xy², showing symmetry in conjugate binomials

Explanation

Expanding each cube reveals that +3x²y and −3x²y cancel. The terms +3xy² from both expansions add together, doubling their coefficient. This creates 2x³ + 6xy², showing symmetry in conjugate binomials

11) (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴.

True

False

Binomial coefficients 1-4-6-4-1 come from Pascal’s triangle for n = 4. Each term follows x^(4−k) y^k with the corresponding coefficient. This ensures all mixed products appear with correct multiplicity.

Explanation

Binomial coefficients 1-4-6-4-1 come from Pascal’s triangle for n = 4. Each term follows x^(4−k) y^k with the corresponding coefficient. This ensures all mixed products appear with correct multiplicity.

12) (a − b)(a² + ab + b²) = a³ − b³.

True

False

Multiplying the binomial and trinomial causes the middle terms to cancel. The result a³ − b³ matches the difference-of-cubes identity exactly. This factorization is essential in algebraic simplification.

Explanation

Multiplying the binomial and trinomial causes the middle terms to cancel. The result a³ − b³ matches the difference-of-cubes identity exactly. This factorization is essential in algebraic simplification.

13) (x + 1)³ − (x − 1)³ = 6x² + 2.

True

False

Apply a³ − b³ = (a − b)(a² + ab + b²) with a = x + 1 and b = x − 1. Evaluate a − b = 2 and a² + ab + b² = 3x² + 1. Multiplying yields 6x² + 2, showing how conjugates simplify cleanly.

Explanation

Apply a³ − b³ = (a − b)(a² + ab + b²) with a = x + 1 and b = x − 1. Evaluate a − b = 2 and a² + ab + b² = 3x² + 1. Multiplying yields 6x² + 2, showing how conjugates simplify cleanly.

14) (a + b)³ − (a − b)³ = 6a²b + 2b³.

True

False

Expanding both cubes shows odd-powered b terms double, while even-powered ones cancel. This produces a simplified expression involving only terms with a²b and b³. Such patterns frequently appear with conjugate expressions.

Explanation

Expanding both cubes shows odd-powered b terms double, while even-powered ones cancel. This produces a simplified expression involving only terms with a²b and b³. Such patterns frequently appear with conjugate expressions.

15) (a + b + c)² = _____

Squaring creates a² + b² + c² plus twice every pair ab, bc, and ca. These cross-terms appear because multiplication distributes across all variables. The identity reflects the complete set of interactions among three terms.

Explanation

Squaring creates a² + b² + c² plus twice every pair ab, bc, and ca. These cross-terms appear because multiplication distributes across all variables. The identity reflects the complete set of interactions among three terms.

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16) (a + b)³ − (a − b)³ = _____

The difference of cubes eliminates even-power cross terms and doubles the odd ones. This produces 6a²b + 2b³ when simplified. It highlights symmetry between conjugate expressions.

Explanation

The difference of cubes eliminates even-power cross terms and doubles the odd ones. This produces 6a²b + 2b³ when simplified. It highlights symmetry between conjugate expressions.

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17) (x − y)⁴ = _____

Using Pascal coefficients 1-4-6-4-1 with alternating signs gives x⁴ − 4x³y + 6x²y² − 4xy³ + y⁴. This pattern is characteristic of even-power binomial expansions.

Explanation

Using Pascal coefficients 1-4-6-4-1 with alternating signs gives x⁴ − 4x³y + 6x²y² − 4xy³ + y⁴. This pattern is characteristic of even-power binomial expansions.

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18) (a + b)² + (a − b)² = _____

Expanding each square reveals +2ab and −2ab, which cancel perfectly. Adding the results leaves 2a² + 2b² with no cross terms. This identity reinforces symmetry in conjugate binomials.

Explanation

Expanding each square reveals +2ab and −2ab, which cancel perfectly. Adding the results leaves 2a² + 2b² with no cross terms. This identity reinforces symmetry in conjugate binomials.

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19) (a + b)³ + (a − b)³ = _____

Adding the cubes removes terms involving odd powers of b and doubles the others. This results in a compact expression involving only a³ and ab². It demonstrates structured cancellation in polynomial identities.

Explanation

Adding the cubes removes terms involving odd powers of b and doubles the others. This results in a compact expression involving only a³ and ab². It demonstrates structured cancellation in polynomial identities.

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20) If a + b = 5 and ab = 6, find a² + b².

13

Use (a + b)² = a² + 2ab + b². Substituting known values gives 25 = a² + b² + 12. Thus a² + b² = 13, showing how identities allow computation without solving for a and b individually.

Explanation

Use (a + b)² = a² + 2ab + b². Substituting known values gives 25 = a² + b² + 12. Thus a² + b² = 13, showing how identities allow computation without solving for a and b individually.