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factorize a³ - 1_a³​

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Answered by Anonymous
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9888628255

07.07.2017

Math

Secondary School

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Factorise : a³ - 1/a³ - 2a + 2/a

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abhi178

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we have to factorise the expression, a³ - 1/a³ - 2a + 2/a

solution : expression, a³ - 1/a³ - 2a + 2/a

we know, x³ - y³ = (x - y)³ + 3xy(x - y)

so, (a³ - 1/a³) = (a - 1/a)³ + 3 × a × 1/a (a - 1/a)

= (a - 1/a)³ + 3(a - 1/a).......(1)

now a³ - 1/a³ - 2a + 2/a = (a³ - 1/a³) - 2(a - 1/a)

= (a - 1/a)³ + 3(a - 1/a) - 2(a - 1/a) [ from eq (1).]

= (a - 1/a)³ + (a - 1/a)

= (a - 1/a)[(a - 1/a)² + 1 ]

= (a - 1/a)[a² + 1/a² - 2 × a × 1/a + 1]

= (a - 1/a)[a² + 1/a² - 2 + 1]

= (a - 1/a)(a² + 1/a² - 1)

Therefore, the factorisation of given expression will be (a - 1/a)(a² + 1/a² - 1)

Answered by amritamohanty918
0

Step-by-step explanation:

(a-1/a) (a2+1/a2-1)

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