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Sagot :
You can simplify it. Or just this one...
1. Cube the first term.
2. Three times the square of the first term and the second term.
3. Three times the first terms and the square of the second term
4. Cube of the second term.
For example:
(a+b)³ = [tex]a^3+3a^2b+3ab^2+b^3[/tex]
(a-b)³ = [tex]a^3-3a^2b+3ab^2-b^3[/tex]
(4b+3)³ = [tex]64b^3+(3)(16b^2)(3)+(3)(4b)(9)+9\\64b^3+144b^2+108b+9[/tex]
1. Cube the first term.
2. Three times the square of the first term and the second term.
3. Three times the first terms and the square of the second term
4. Cube of the second term.
For example:
(a+b)³ = [tex]a^3+3a^2b+3ab^2+b^3[/tex]
(a-b)³ = [tex]a^3-3a^2b+3ab^2-b^3[/tex]
(4b+3)³ = [tex]64b^3+(3)(16b^2)(3)+(3)(4b)(9)+9\\64b^3+144b^2+108b+9[/tex]
Cube of binomial pattern, where a = first term and b = second/last term:
(a + b)³ = (a + b) (a + b) (a + b)
= a³ + 3a²b + 3ab² + b³
Steps in finding the special product of (a + b)³ :
a) Cube the first term, a = a³
b) Three (first term)² (last term) = 3 (a)² (b) = 3a²b
c) Three (first term) (last term)² = 3 (a) (b)² = 3ab²
d) Cube the last term = b³
Combine the results, the special product is:
a³ + 3a²b + 3ab³ + b³
(a + b)³ = (a + b) (a + b) (a + b)
= a³ + 3a²b + 3ab² + b³
Steps in finding the special product of (a + b)³ :
a) Cube the first term, a = a³
b) Three (first term)² (last term) = 3 (a)² (b) = 3a²b
c) Three (first term) (last term)² = 3 (a) (b)² = 3ab²
d) Cube the last term = b³
Combine the results, the special product is:
a³ + 3a²b + 3ab³ + b³
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