Question

Resolve resolve X³/(x-a){x-b}[x-c] into partial fractions please​

Answer 1

Answer:

Resolve x^3/((x-a)(x-b)(x-c)) into partial fractions.

Updated On: 22-2-2020

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Answer 2

Step-by-step explanation:

Given :-

X^3/(x-a){x-b}[x-c)

To find :-

Resolve r X^3/(x-a){x-b}[x-c] into partial fractions?

Solution :-

Given that

x^3/(x-a){x-b}[x-c)

Let x^3/(x-a){x-b}[x-c) = A/(x-a)+ B/(x-b) + C/(x-c)

x^3/(x-a){x-b}[x-c) = >

[A(x-b)(x-c)+ B(x-a)(x-c) + C(x-a)(x-b)]/(x-a){x-b}[x-c)

On cancelling (x-a)(x-b)(x-c) both sides then

x^3 = A(x-b)(x-c)+ B(x-a)(x-c) + C(x-a)(x-b)-----(1)

Put x = a in (1) then

a^3 = A(a-b)(a-c)+ B(a-a)(a-c) + C(a-a)(a-b)

=>a^3 = A(a-b)(a-c)+B(0)+C(0)

=> a^3 = A(a-b)(a-c)+0+0

=> a^3 = A(a-b)(a-c)

=>A = a^3/(a-b)(a-c)--------(2)

Put x = b in (1) then

=> b^3 = A(b-b)(b-c)+ B(b-a)(b-c) + C(b-a)(b-b)

=> b^3 = A(0)+B(b-a)(b-c)+C(0)

=> b^3 = 0+B(b-a)(b-c)+0

=> b^3 = B(b-a)(b-c)

=> B = b^3/(b-a)(b-c) ---------(3)

Put x = c in (1) then

c^3 = A(c-b)(c-c)+ B(c-a)(c-c) + C(c-a)(c-b)

=> c^3 = A(0)+B(0)+C(c-a)(c-b)

=> c^3 = 0+0+C(c-a)(c-b)

=> c^3 = C(c-a)(c-b)

=> C = c^3/(c-a)(c-b)---------(4)

Now

On Substituting the values of A ,B and C in the given equation then it becomes

x^3/(x-a){x-b}[x-c) = A/(x-a)+ B/(x-b) + C/(x-c)

=> [a^3/(a-b)(a-c)]/(x-a)+[ b^3/(b-a)(b-c)]/(x-b)

+[c^3/(c-a)(c-b)/(x-c)

Answer:-

x^3/(x-a){x-b}[x-c) = [a^3/(a-b)(a-c)]/(x-a)+[b^3/(b-a)(b-c)]/(x-b) +[c^3/(c-a)(c-b)/(x-c)