Question

Factorise a4-a^3+a-1

Answer 1

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dude here is your answer⏭️⏭️

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a4-a³+a-1.

(a4-a3)+(a-1) taking common

a³(a-1)+1(a-1)

(a³+1) (a-1)

Answer 2

split the expression at hand into groups, each group having two terms :

Group 1: -a+1

Group 2: a4-a3

Pull out from each group separately :

Group 1: (-a+1) • (1) = (a-1) • (-1)

Group 2: (a-1) • (a3)

Add up the two groups :

(a-1) • (a3-1)

Factoring: a3-1

Theory : A difference of two perfect cubes, a3 - b3 can 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 : 1 is the cube of 1

Check : a3 is the cube of a1

Factorization is :

(a - 1) • (a2 + a + 1)

Factoring a2 + a + 1

The first term is, a2 its coefficient is 1 .

The middle term is, +a its coefficient is 1 .

The last term, "the constant", is +1

Step-1 : Multiply the coefficient of the first term by the constant 1 • 1 = 1

Step-2 : Find two factors of 1 whose sum equals the coefficient of the middle term, which is 1 .

-1 + -1 = -2

1 + 1 = 2

Multiply (a-1) by (a-1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (a-1) and the exponents are :

1 , as (a-1) is the same number as (a-1)1

and 1 , as (a-1) is the same number as (a-1)1

The product is therefore, (a-1)(1+1) = (a-1)2

Final result :

(a - 1)2 • (a2 + a + 1)