Factorize each of the following expressions:
a³ - 1/a³ -2a 2/a
Answers
The algebraic expression is given as : a³ - 1/a³ - 2a + 2/a
= a³ - 1/a³ - 2(a - 1/a)
In order to factorise the given algebraic expression as the difference of two cubes we use the following identities - a³ - b³ = (a - b) (a² + ab + b²) :
= (a - 1/a) (a² + 1/a² + a × 1/a ) - 2(a - 1/a)
= (a - 1/a) (a² + 1/a² + 1 ) - 2(a - 1/a)
Here (a - 1/a) is a common factor :
= (a - 1/a) [(a² + 1/a² + 1 - 2)]
= (a - 1/a) [(a² + 1/a² – 1 )]
a³ - 1/a³ - 2a + 2/a = (a - 1/a) [(a² + 1/a² – 1 )]
Hence, the factorisation of a³ - 1/a³ - 2a + 2/a is (a - 1/a) [(a² + 1/a² – 1 )]
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Answer:
Step-by-step explanation:
= (a - 1/a) (a² + 1/a² + a × 1/a ) - 2(a - 1/a)
= (a - 1/a) (a² + 1/a² + 1 ) - 2(a - 1/a)
Here (a - 1/a) is a common factor :
= (a - 1/a) [(a² + 1/a² + 1 - 2)]
= (a - 1/a) [(a² + 1/a² – 1 )]
a³ - 1/a³ - 2a + 2/a = (a - 1/a) [(a² + 1/a² – 1 )]