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Question : 49 of 86

Marks: +1, -0

If

a x^{\frac{2}{3}} + b x^{\frac{1}{3}} + c = 0,

then the value of

a^{3} x^{2} +

b^{3} x + c^{3}

$3 a b c x$
$-3 a b c x$
$3 a b c$
$-3 a b c$ .

Solution:

Concept:

Substitute

y = x^{1/3}

to transform the equation into a quadratic, then use the identity

p^3+q^3+r^3 = 3pqr

when

p+q+r=0

.

Explanation:

Let

y = x^{\frac{1}{3}}

, so

x^{\frac{2}{3}} = y^2

and

x = y^3

,

x^2 = y^6

.

The given equation becomes

a y^2 + b y + c = 0

.

Define

p = a y^2

,

q = b y

,

r = c

. Then

p+q+r = 0

.

We need

a^3 x^2 + b^3 x + c^3 = a^3 y^6 + b^3 y^3 + c^3 = p^3 + q^3 + r^3

.

Using the identity

p^3+q^3+r^3 = 3pqr

when

p+q+r=0

, we get:

p^3+q^3+r^3 = 3 \cdot (a y^2) \cdot (b y) \cdot c = 3 a b c y^3 = 3 a b c x

.

Answer:

3 a b c x

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