Question

$$\frac{15}{16}y^3 + 4 - \frac{1}{8}y^3 = square$$ (use the operation symbols in the math pal

Explanation:

Step1: Combine like terms

We have \(\frac{15}{16}y^{3}\) and \(-\frac{1}{8}y^{3}\). First, convert \(-\frac{1}{8}y^{3}\) to sixteenths: \(-\frac{1}{8}y^{3}=-\frac{2}{16}y^{3}\).

Step2: Add the coefficients of \(y^{3}\)

Now, add \(\frac{15}{16}y^{3}\) and \(-\frac{2}{16}y^{3}\): \(\frac{15}{16}y^{3}-\frac{2}{16}y^{3}=\frac{15 - 2}{16}y^{3}=\frac{13}{16}y^{3}\).

Step3: Include the constant term

The constant term is \(4\), so the simplified expression is \(\frac{13}{16}y^{3}+4\).

Answer:

\(\frac{13}{16}y^{3}+4\)