Question

Factorise: x(b+c-a) + x(c+a-b) + x(a+b-c)​

Answer 1

Answer:

If (x - a)/(b+c) + (x - b)/(c + a) + (x - c)/(a + b) = 3, then what is the value of x?

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Subtract 3 from both sides but creatively:

x−ab+c−1+x−ba+c−1+x−ca+b−1=0

x−a−b−cb+c+x−a−b−ca+c+x−a−b−ca+b=0

Taking common:

(x−a−b−c)×k=0

Where k is some random expression in a,b, and c.

This implies,

x−a−b−c=0

x=a+b+c

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given

(x - a)/(b+c) + (x - b)/(c + a) + (x - c)/(a + b) = 3

now take right hand side

,(x - a)/(b+c) + (x - b)/(c + a) + (x - c)/(a + b)

just for sake to convert numerator in denominator variable let x= a+b+c

thus (x-a)=b+c,(x-b)=c+a,(x-c)=a+b

thats why

(b+c)/(b+c) + (c+a)/(c + a) + (a+b)/(a + b) = 3 =RHS

it means the value of x will be (a+b+c) because thats what we inserted it when we started solving the question..

all such questions are solved by letting a value either for x or y..

Answer 2

Answer:

x(b+c+a)

Step-by-step explanation:

bx+cx-ax+cx+ax-bx+ax+bx-cx

bx+cx+ax

or, x(b+c+a)

I hope it help you...

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