Question

Factorise a^{​​​​​​6}​​​​​-b^​​​6

Answer 1

a⁶-b⁶ Final result : (a+b)•(a²-ab+b²)•(a-b)•(a²+ab+b²)

Step by step solution : Step1: Trying to factor as a Difference of Squares:

1.1 Factoring:a⁶-b⁶ Theory : A difference of two perfect squares,A²-B²can be factored into(A+B)•(A-B) Proof:(A+B)•(A-B)= A²-AB+BA-B²= A²-AB+ AB- B²= A²- B² Note :AB = BAis the commutative property of multiplication. Note :-AB+ ABequals zero and is therefore eliminated from the expression. Check:a⁶is the square ofa³ Check:b⁶is the square ofb³ Factorization is :(a³+ b³)•(a³- b³)

Trying to factor as a Sum of Cubes:

1.2 Factoring:a³+ b³ Theory:A sum of two perfect cubes,a³+b³can be factored into : (a+b)•(a²-ab+b²) Proof: (a+b)•(a²-ab+b²) = a³-a²b+ab²+ba²-b²a+b³= a³+(a²b-ba²)+(ab²-b²a)+b³= a³+0+0+b³= a³+b³ Check: a³is the cube of a¹ Check: b³is the cube of b¹ Factorization is : (a + b)•(a²- ab + b²)

Trying to factor a multi variable polynomial :

1.3 Factoringa²- ab + b² Try to factor this multi-variable trinomial using trial and error Factorization fails

Trying to factor as a Difference of Cubes:

1.4 Factoring:a³- b³ Theory : A difference of two perfect cubes,a³-b³can be factored into (a-b)•(a²+ab+b²) Proof:(a-b)•(a²+ab+b²)= a³+a²b+ab²-ba²-b²a-b³= a³+(a²b-ba²)+(ab²-b²a)-b³= a³+0+0+b³= a³+b³ Check: a³is the cube of a¹ Check: b³is the cube of b¹ Factorization is : (a - b)•(a²+ ab + b²)

Trying to factor a multi variable polynomial :

1.5 Factoringa²+ ab + b² Try to factor this multi-variable trinomial using trial and error Factorization fails

Final result : (a+b)•(a²-ab+b²)•(a-b)•(a²+ab+b²)