Question

how to solve question number 154?

Answer 1

Hi friend, Harish here.

Here is your answer:

Let ∛a = x ,  ∛b = y , ∛c = z.

Given that, x + y + z =0.

We know that, 

(x³ + y³ + z³) - 3xyz = (x+y+z) (x²+y²+z²-(xy+yz+zx))

Now, substitute x + y + z value here.

Then,

⇒ (x³ + y³ + z³) - 3xyz = (0)(x²+y²+z²-(xy+yz+zx)) = 0.

So, x³ + y³ + z³ = 3xyz.   ---- (i)

We know that,

$(\frac{a+b+c}{3})^{3} = (\frac{x^{3}+y^{3}+z^{3}}{3})^{3}$

Now, substitute (i) in here.

Then,

⇒ $(\frac{x^{3}+y^{3}+z^{3}}{3})^{3} = (\frac{3xyz}{3})^{3} = (xyz)^{3]$

⇒ $(xyz)^{3}= (\sqrt[3]{a\times b\times c})^{3} = abc.$

Therefore the answer is abc.(OPTION - A)
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Hope my answer is helpful to you.

Answer 2

we have given (a)1/3+(b)^1/3+(c)^1/3=0
also( (a)^1/3+(b)^1/3+(c)^1/3)^3=0
((a)^1/3+(b)^1/3)^3+c+3((a)^1/3+(b)^1/3)(c)^1/3=0 (a^1/3+b^1/3+c^1/3)=0
(a^1/3+b^1/3)^3+c =0
a+b+c+3(a)^1/3 (b)^1/3 ((a)^1/3+(b)^1/3)=0
a+b+c= -3(a)^1/3(b)^1/3 (-(c)^1/3
a+b+c=3 (a)^1/3 (b)^1/3 (c)^1/3
now put it in the eq
{(3 (a)^1/3 (b)^1/3 (c)^1/3)/3}^3
simplifying this we get
ans= abc..