## Math 8 Chapter 1 Lesson 5: Memorable Equivalence Constants (continued)

## 1. Theoretical Summary

### 1.1. Sum of two cubes

\({A^3} + {B^3} = (A + B)({A^2} – AB + {B^2})\)

### 1.2. Difference of two cubes

\({A^3} – {B^3} = (A – B)({A^2} + AB + {B^2})\)

Like the previous equality constants, the proof of these two equality constants is also based on the rule Multiplying polynomials by polynomials that we have learned.

## 2. Illustrated exercise

**Question 1:** Rewrite the following expression as sum or difference:

\(\left( {2x – 4{y^2}} \right)\left( {4{x^2} + 8x{y^2} + 16{y^4}} \right)\)

**Solution guide**

\(\begin{array}{l} \left( {2x – 4{y^2}} \right)\left( {4{x^2} + 8x{y^2} + 16{y^4} } \right)\\ = \left( {2x – 4{y^2}} \right)\left[ {{{\left( {2x} \right)}^2} + \left( {2x} \right)\left( {4{y^2}} \right) + {{\left( {4{y^2}} \right)}^2}} \right]\\ = {\left( {2x} \right)^3}-{\left( {4{y^2}} \right)^3}\\ = 8{x^3} – 64{y^6 } \end{array}\)

**Verse 2:** Prove that::\(({x^3} + {y^3})({x^3} – {y^3}) = ({x^2} – {y^2})({x^) 4} + {x^2}{y^2} + {y^4})\)

**Solution guide**

We have the left side: \(({x^3} + {y^3})({x^3} – {y^3}) = {\left( {{x^3}} \right)^2 } – {\left( {{y^3}} \right)^2} = {x^6} – {y^6}\)

We have the right side:

\(\begin{array}{l} ({x^2} – {y^2})({x^4} + {x^2}{y^2} + {y^4})\\ = ({x^2} – {y^2})\left[ {{{\left( {{x^2}} \right)}^2} + {x^2}{y^2} + {{\left( {{y^2}} \right)}^2}} \right]\\ = {\left( {{x^2}} \right)^3} – {\left( {{y^2}} \right)^3}\\ = {x^6} – {y^ 6}\\ \end{array}\)

Notice that the left side is equal to the right side.

So we have something to prove.

**Question 3:** Prove that: \(({11^3} – 1)^n\) is divisible by \(10^n\).

**Solution guide**

We can transform the expression as follows:

\(\begin{array}{l} {\left( {{{11}^3} – 1} \right)^n}\\ = {\left[ {(11 – 1)({{11}^2} + 11 + 1)} \right]^n}\\ = {\left[ {(10)({{11}^2} + 11 + 1)} \right]^n}\\ = {10^n}{({11^2} + 11 + 1)^n} \end{array}\)

notice that this expression is divisible by \(10^n\). So we have something to prove.

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Simplify the following expressions:

a) \(\left( x+2 \right)\left( {{x}^{3}}-2x+4 \right)-\left( 24+{{x}^{3}} \right) \)

b) \(\left( 3x-y \right)\left( 9{{x}^{2}}+3xy+{{y}^{2}} \right)-\left( 3x+y \right) \left( 9{{x}^{2}}-3xy+{{y}^{2}} \right)\)

**Verse 2: **Rewrite the following expression as sum or difference: \(\left( x-3{{y}^{2}} \right)\left( {{x}^{2}}+3x{{y}^) {2}}+9{{y}^{4}} \right)\)

### 3.2. Multiple choice exercises

**Question 1: **Which of the following is the sum of two cubes?

A. \(A^3+B^3=(A+B)(A^2-AB+B^2)\)

B. \(A^3+B^3=(AB)(A^2+AB+B^2)\)

C. \((A+B)^3=A^3+3A^2B+3AB^2+B^3\)

D. \((AB)^3=A^3-3A^2B+3AB^2-B^3\)

**Verse 2: **Expanding the equality of two cubes gives us which of the following results?

A. \((AB)^3=A^3-3A^2B+3AB^2-B^3\)

B. \(A^3-B^3=(AB)(A^2+AB+B^2)\)

C. \(A^3-B^3=(A+B)(A^2-AB+B^2)\)

D. \(A^3-B^3=(AB)(A+B)\)

**Question 3: **The value of the formula \(P=8x^3+12x^2+6x+1\) at \(x=2\) is ?

A. 27

B. 48

C. 8

D. 75

**Question 4: **Write the following expression as the cube of a difference

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

A. \({\left( {x – \frac{1}{2}} \right)^2}\)

B. \((2x-1)^2\)

C. \((x-3)^3\)

D. \({\left( {x – \frac{1}{2}} \right)^3}\)

**Question 5: **Expanding \({\left( {a + b + c} \right)^3}\) gives us which of the following results?

A. \({\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3\left( {a + b} \ right)\left( {b + c} \right)\left( {c + a} \right)\)

B. \({\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3ab(b + c)\)

C. \({\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3abc\left( {a + b + c) } \right)\)

D. \({\left( {a + b + c} \right)^3} = {a^3} + {b^3} + {c^3} + 3ab + 3bc + 3ca\)

## 4. Conclusion

Through this lesson, you should achieve the following goals:

- Remember the equality of the sum of the cubes and the difference of two cubes.
- Apply the learned equality constants to solve related problems.

.

=============

## Leave a Reply