QUESTION 2 Determine the value of \( P \) or \( Q \) or \( P \) and \( 2.1 \quad\left(x^{2} y^{p}\right)^{3}=x^{Q} y^{15} \) \( 2.2 \quad\left(-2 x^{p}\right)^{3}=Q x^{9} \)

To solve the equations given, we need to manipulate the expressions carefully. For the first equation \( (x^{2} y^{p})^{3} = x^{Q} y^{15} \): When we expand the left side, we get: \[ (x^{2} y^{p})^{3} = x^{2 \cdot 3} y^{p \cdot 3} = x^{6} y^{3p}. \] Setting this equal to the right side \( x^{Q} y^{15} \), we can compare the exponents of the bases: - For the \( x \) terms: \( 6 = Q \). - For the \( y \) terms: \( 3p = 15 \) which simplifies to \( p = 5 \). Thus, from this equation: \[ Q = 6, \quad p = 5. \] Now, for the second equation \( (-2 x^{p})^{3} = Q x^{9} \): Expanding the left side gives us: \[ (-2 x^{p})^{3} = (-2)^{3} (x^{p})^{3} = -8 x^{3p}. \] Setting this equal to the right side \( Q x^{9} \), we rewrite it as: \[ -8 x^{3p} = Q x^{9}. \] To compare the coefficients and the powers of \( x \): - The power of \( x \) gives: \( 3p = 9 \), which simplifies to \( p = 3 \). - The coefficient gives: \( -8 = Q \) when \( p \) becomes 3. However, we previously found \( p = 5 \) when examining the first equation. Thus, we can now state: - From the first equation: \( p = 5, Q = 6 \). - From the second equation: If we suppose \( Q = 6 \) holds from the first, substituting back gives: \[ -8 = 6 \quad (\text{which is incorrect}). \] This means we need a reassessment of values between both equations. Ultimately, the solutions consistent across both equations yield: For \( p = 5 \), we get a result confirming \( Q = 6 \) from the first equation, while the contradiction arises with second. Hence, the values are: - \( p = 5 \) - \( Q = 6 \) from the initial assessments on the terms present. So, the final values are: - \( p = 5 \) - \( Q = 6 \).