AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the equation.
‌‌

x4

(x−a)(x−b)(x−c)

=P(x)+‌‌

A

x−a

+‌‌

B

x−b

+‌‌

C

x−c

The above equation is solved as,
x4=[

P(x)(x−a)(x−b)(x−c)+A(x−b)(x−c)+

B(x−a)(x−c)+C(x−a)(x−b)

]
Substitute 0 for x in the above expression.
P(0)abc=[Abc+Bac+Cab]
P(0)=‌‌

A

a

+‌‌

B

b

+‌‌

C

c

Now at x=a
A=‌‌

a4

(a−b)(a−c)

At x=b
B=‌‌

b4

(b−c)(b−a)

At x=c
C=‌‌

c4

(c−a)(c−b)

Thus
P(0)=‌‌

a3

(a−b)(a−c)

+‌‌

b3

(b−c)(b−a)

+‌‌

c3

(c−a)(c−a)

=‌‌

a3(c−b)+bc(c2−b2)+a(b3−c3)

(a−b)(b−c)(c−a)

=‌‌

a3+bc(c+b)−a(b2+c2+bc)

(a−b)(a−c)

=‌‌

a(a2−b2)−c2(a−b)−bc(a−b)

(a−b)(a−c)

Solve further
P(0)=‌‌

(a2+ba)−c2−bc

(a−c)

=a + b + c
Therefore,
P(0)+(a−b)(a−c)A=a+b+c+a4