If x is a positive real number such that x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4, what is t… — GMAT

If $x$ is a positive real number such that $x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4$ , what is the value of $x^{\frac{3}{2}} + x^{-\frac{3}{2}}$ ?

Cubing both sides of

x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4

yields

(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^3 = 64

. By the identity

(a+b)^3 = a^3 + b^3 + 3ab(a+b)

, the left side expands to

(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}})(x^{\frac{1}{2}} + x^{-\frac{1}{2}})

. Substituting

x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1

and

x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4

gives

(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 = 64

. Subtracting 12 yields 52.

Set up the cubic identity for the sum of fractional exponents.

Let

a = x^{\frac{1}{2}}

and

b = x^{-\frac{1}{2}}

. Then

a + b = 4

and

ab = x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1

Recognizing that

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

and

x^{-\frac{3}{2}} = b^3

allows the use of binomial expansion.

Cube both sides of the given equation

a + b = 4

(a + b)^3 = a^3 + b^3 + 3ab(a + b) \implies 4^3 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(1)(4)

Expanding the cube retains the desired expression

a^3 + b^3

alongside simpler terms.

Simplify and solve for

x^{\frac{3}{2}} + x^{-\frac{3}{2}}

64 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 \implies x^{\frac{3}{2}} + x^{-\frac{3}{2}} = 64 - 12 = 52.

Subtracting 12 from 64 gives the exact value of the expression.

Algebraic identities with fractional exponents

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