1 + b + bc + b^c factorize this expression
Answers
Answer:
=(a^2+b^2) /ab+(b^2+c^2) /bc+(c^2+a^2) /ca+3
=(ca^2+cb^2+ab^2+ac^2+bc^2+ba^2+3abc) /abc
={a^2(c+b+a-a) +b^2(c+a+b-b) +c^2(a+b+c-c) +3abc}/abc
={(a+b+c) (a^2+b^2+c^2) -(a^3+b^3+c^3–3abc) }/abc
={(a+b+c) (a^2+b^2+c^2-a^2-b^2-c^2+ab+bc+ca) })/abc
={(a+b+c) (ab+bc+ca) }/abc
=(a+b+c) (1/c+1/a+1/b)
=(a+b+c) (1/a+1/b+1/c)…… read factorisation
x+y+z=89/60, which is b/(a+b)+c/(b+c)+a/(c+a) exactly.
So if a, b, c are not 0, then the answer is 89/60.
If at least one of a, b, c is 0, say a=0.
Then we have (b-c)/(b+c)=1/30, b=c*31/29.
The answer is 89/60.
So there is only one solution: 89/60.
Rajendra Rajput
Answered 4 years ago
If ab+bc+ca=0 , find the value of 1a2−bc+1b2−ca+1c2−ab ?
ab+bc+ca=0
ab+bc = -ca
Or ab+ca= -bc
Or bc+ca = -ab
Now
1/(a^2 -bc) + 1/(b^2 -ca) + 1/(c^2 -ab)
= 1/(a^2 +ab+ca) +1/(b^2 +ab+bc) +1/(c^2 +bc+ca)
= 1/(a+b+c) * (1/a + 1/b + 1/c)
= 1/(a+b+c) * (bc + ca + ab)/abc
= (ab+bc+ca)/(a+b+c) abc
= (a+b+c) abc/(a+b+c) abc
= 0
Kumar Saurav
Nirvair Atwal
Answered 3 years ago
What is the answer of if 4/3 = a/b then a+b=?
Jayanta Mukherjee
Answered 1 year ago
How do you factorize 1+a+b+c+ab+bc+CA+abc?
1 + a + b + c + ab + bc + ca + abc
Just rejig things
= 1 + a + b + ab + c + ca + bc + abc
= 1*(1 + a) + b*(1 + a) + c*(1 + a) + bc*(1 + a)
= (1 + a) * (1 + b + c + bc)
= (1 + a) * {1*(1 + b) + c*(1 + b)}
= (1 + a) * (1 + b) * (1 + c).
Ruchi Chhabra
Thirisha Mandalapu
Answered 2 years ago
How do I factorize a^3 + b^3 + a + b?
Given a^3 +b^3 +a + b
a3 + b3 = (a + b) (a^2 - ab + b^2 ).
= (a + b) (a^2 - ab + b^2 )+(a+b)
=(a+b)(a^2 + ab +b^2 +1).
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