Math, asked by rkcomp31, 4 months ago

If a³+b³+c³=3abc ,then prove that a+b+c=0.


adarsharyan46: well i answered

Answers

Answered by krishpatel2383
2

Answer:

From the given equation a^3+b^3+c^3=3abc

Here by using the algebraic identity :

a^3+b^3+c^3-3abc=(a+b+c) (a^2+b^2+c^2-ab-bc-ac)

Substitute the values of the formula

(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0

(a+b+c)=0

Step-by-step explanation:

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adarsharyan46: well right
adarsharyan46: not perfect tho
Answered by adarsharyan46
1

Step-by-step explanation:

Firstly,

a^{3} +b^{3} +c^{3} = 3abc

take 3abc to LHS

a^{3} +b^{3} +c^{3} - 3abc = 0

Now,

[ we know that a^{3} +b^{3} +c^{3} - 3abc = (a +b +c)(a^{2} b^{2} c^{2} -ab -bc -ac)]

hence,

(a +b +c)(a^{2} b^{2} c^{2} -ab -bc -ac)  = 0

Now we take (a +b +c)(a^{2} b^{2} c^{2} -ab -bc -ac) to RHS so it multiplies with 0

giving us,

(a+b+c)=0

Hence proved..


adarsharyan46: how bout this?
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