Math, asked by rani712005, 10 months ago

Tomorrow is my exam...
Please help!!!!

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rani712005: The question is aa in the photo

rani712005: As

DrStrange1224: u can multiply throughout the equation with any power.....in this question u had to eliminate 3 as the denominator

DrStrange1224: hence took a whole power of 3 throughout the equation

DrStrange1224: a,b,c are variables

DrStrange1224: so u cant take any values for a ,b and c

DrStrange1224: hence u have to eliminate that denominator to obtain the equation a+b+c

rani712005: Please help yaar

rani712005: Write in answee boc

rani712005: Box

Answers

Answered by DrStrange1224

$({a}^{ \frac{1}{3}})^{3} + ({b}^{ \frac{1}{3}})^{3} + ({c}^{ \frac{1}{3}})^{3} = (0) ^{3}$

$a + b + c = 0$

Anonymous: 'Fraid that's not right. This is using x^3+y^3+z^3 = (x+y+z)^3... which is NOT true! To help see that the conclusion is wrong, put a = -1, b = -8, c = 27. Then a^(1/3) + b^(1/3) + c^(1/3) = -1 -2+3 = 0, as required, but a+b+c=18... not 0 !!

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Tomorrow is my exam...
Please help!!!!