What happens when we replace b with −b in this new identity?

Step-by-step solution

Idea: “This new identity” is (a + b)³ = a³ + 3a²b + 3ab² + b³. Because it holds for every value of b, it also holds with −b in place of b. Then use: an even power of −b is positive, an odd power is negative.

1. [a + (−b)]³ = a³ + 3a²(−b) + 3a(−b)² + (−b)³½ mark
2. Simplify: 3a²(−b) = −3a²b; (−b)² = b²; (−b)³ = −b³.½ mark
3. So (a − b)³ = a³ − 3a²b + 3ab² − b³.½ mark
4. Of the four terms, two are positive and two are negative, and they alternate.½ mark

Check: a = 5, b = 2: 3³ = 27 and 125 − 150 + 60 − 8 = 27 ✓.

Answer to write in the exam

Common mistakes that cost marks

• Making every sign negative: a³ − 3a²b − 3ab² − b³. The third term is + because (−b)² = b².
• Writing (a − b)³ = a³ − b³.
• Writing (−b)³ = b³. An odd power of a negative is negative.

How this can come in the exam

Try one yourself

Expand (2p − 3)³.

More questions like this

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• Now consider the expression 8n³ − 60n²m + 150nm² − 125m³. If you write it in the form (a − b)³, what will be a and b?
• Now let us play with known identities to discover more identities. Try to multiply the following using the distributive property.
• We already know that x² − y² = (x − y)(x + y).
  Further, we have verified that x³ − y³ = (x − y)(x² + xy + y²).
  Observe that x − y is a common factor of x² − y² and x³ − y³.
  Do you think x − y is also a factor of x⁴ − y⁴?
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  Can you see how x − y is a factor of x⁴ − y⁴?
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• Exploring further, let us multiply (x + y + z) and (x² + y² + z² − xy − xz − yz).

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