\( P = \frac { a ^ { \frac { 6 } { 5 } } b + b ^ { \frac { 6 } { 5 } } a } { \sqrt[ 5 ] { a } + \sqrt[ 5 ] { b } } ( a , b > 0 ) \)

The given expression is: \[ P = \frac{a^{\frac{6}{5}} b + b^{\frac{6}{5}} a}{\sqrt[5]{a} + \sqrt[5]{b}} \quad \text{where} \quad a, b > 0 \] Let's simplify \( P \). ### Step 1: Express in Terms of Exponents Rewrite the expression using exponent notation: \[ P = \frac{a^{6/5} \cdot b + b^{6/5} \cdot a}{a^{1/5} + b^{1/5}} \] ### Step 2: Substitute Variables Let’s make the substitution to simplify the expression: \[ x = a^{1/5} \quad \text{and} \quad y = b^{1/5} \] Then: \[ a = x^5 \quad \text{and} \quad b = y^5 \] Substituting these into \( P \): \[ P = \frac{(x^5)^{6/5} \cdot y^5 + (y^5)^{6/5} \cdot x^5}{x + y} = \frac{x^6 y^5 + y^6 x^5}{x + y} \] ### Step 3: Factor the Numerator Notice that the numerator can be factored: \[ x^6 y^5 + y^6 x^5 = x^5 y^5 (x + y) \] Thus: \[ P = \frac{x^5 y^5 (x + y)}{x + y} = x^5 y^5 \] ### Step 4: Re-substitute Original Variables Recall that: \[ x^5 = a \quad \text{and} \quad y^5 = b \] Therefore: \[ P = a b \] ### Verification To ensure the simplification is correct, let's verify with specific values: **Example 1:** - Let \( a = 1 \) and \( b = 1 \): \[ P = \frac{1^{6/5} \cdot 1 + 1^{6/5} \cdot 1}{1^{1/5} + 1^{1/5}} = \frac{1 + 1}{1 + 1} = \frac{2}{2} = 1 \] \[ ab = 1 \cdot 1 = 1 \quad \checkmark \] **Example 2:** - Let \( a = 32 \) and \( b = 1 \) (since \( 32 = 2^5 \)): \[ P = \frac{32^{6/5} \cdot 1 + 1^{6/5} \cdot 32}{32^{1/5} + 1^{1/5}} = \frac{64 + 32}{2 + 1} = \frac{96}{3} = 32 \] \[ ab = 32 \cdot 1 = 32 \quad \checkmark \] Both examples confirm that \( P = ab \). ### **Final Simplified Form** \[ P = a \cdot b \]