Math, asked by abhi358, 1 year ago

the 42nd question if a b c are all non zero and a+b+ =0...

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8290geet: {abc(a^3+b^3+ c^3)}/(abc×abc)

8290geet: abc and abc cancel out

8290geet: (a^3 +b^3 +c^3)/abc

8290geet: now a^3+b^3+c^3= (a+b+c)^3 -3(a+b)(b+c)(c+a)

8290geet: {(a+b+c)^3 -3(a+b)*(b+c)(c+a)}/abc

8290geet: {-3(-c)(-a)(-b)}/abc

8290geet: 3abc/abc

8290geet: 3 is the answer hp

abhi358: thnx mate

8290geet: welcome

Answers

Answered by sahuraj457

$a + b + c = 0 \\ \frac{ {a}^{2} }{ {bc}^{2} } + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} = \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} = \frac{3abc}{abc} = 3 \\$
if a+b+c= 0 then a3 +b3+C3 = 3abc

abhi358: how did denominator become abc

8290geet: it's wrong mathod

abhi358: 8290geet so pls can u tell me i read ur profile ur preparing for jee u can do these easily

8290geet: yes

abhi358: so...u can pls tell me how to do it?

8290geet: i have this question's answer but i can't

8290geet: give this

Answered by 8290geet

is the answer of the question

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abhi358: thnx now it is much more clear

8290geet: welcome

abhi358: hey man i need ur help

abhi358: please one question its really tough

abhi358: x^3-1/x^3 -14

abhi358: factorise

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