Show that \( 27^{y+1}+3^{3 y} \) can be written in the form \( a\left(b^{\nu}\right) \), where \( a \) and \( b \) are constants. Hence, find the values of \( a \) and of \( b \).

To express \( 27^{y+1} + 3^{3y} \) in the form \( a(b^{\nu}) \), we start by rewriting the terms in the expression using powers of 3. First, we note that: \[ 27 = 3^3 \] Thus, we can rewrite \( 27^{y+1} \) as: \[ 27^{y+1} = (3^3)^{y+1} = 3^{3(y+1)} = 3^{3y + 3} \] Next, we rewrite \( 3^{3y} \) as it is: \[ 3^{3y} = 3^{3y} \] Now we can combine these two terms: \[ 27^{y+1} + 3^{3y} = 3^{3y + 3} + 3^{3y} \] Next, we factor out the common term \( 3^{3y} \): \[ 3^{3y + 3} + 3^{3y} = 3^{3y}(3^3 + 1) \] Calculating \( 3^3 + 1 \): \[ 3^3 = 27 \quad \text{so} \quad 3^3 + 1 = 27 + 1 = 28 \] Thus, we have: \[ 27^{y+1} + 3^{3y} = 3^{3y} \cdot 28 \] Now we can express this in the form \( a(b^{\nu}) \): \[ 27^{y+1} + 3^{3y} = 28 \cdot 3^{3y} \] From this expression, we can identify the constants \( a \) and \( b \): - \( a = 28 \) - \( b = 3 \) Thus, the values of \( a \) and \( b \) are: \[ \boxed{28} \quad \text{and} \quad \boxed{3} \]