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Please find the answer to your question below
Given ab+bc+ca=0 and asked to find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc)
Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac
you get :
=(1/(a2+ab+ac))+(1/(c2+ac+bc))+(1/(a2+ab+ac))
=(1/a(a+b+c))+(1/b(a+b+c))+(1/c(a+b+c))
By taking LCM we get
=(ab + bc + ca)/[(a+b+c)(abc)]
put, ab+bc+ca=0
you get value = 0/[(a+b+c)(abc)] = 0
HOPE IT WILL HELP YOU
PLEASE MARK IT AS BRAINLIEST ANSWER
Given ab+bc+ca=0 and asked to find value 1/(a2-bc ) + 1/(c2-ab) + 1/(a2-bc)
Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + ac
you get :
=(1/(a2+ab+ac))+(1/(c2+ac+bc))+(1/(a2+ab+ac))
=(1/a(a+b+c))+(1/b(a+b+c))+(1/c(a+b+c))
By taking LCM we get
=(ab + bc + ca)/[(a+b+c)(abc)]
put, ab+bc+ca=0
you get value = 0/[(a+b+c)(abc)] = 0
HOPE IT WILL HELP YOU
PLEASE MARK IT AS BRAINLIEST ANSWER
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