Solve for a d=-f^3+1/4ta

$d = - f^{3} + \frac{1}{4} t a$

Rewrite the equation as $- f^{3} + \frac{1}{4} \cdot (t a) = d$ .

$- f^{3} + \frac{1}{4} \cdot (t a) = d$

Multiply $\frac{1}{4} (t a)$ .

Combine $t$ and $\frac{1}{4}$ .

$- f^{3} + \frac{t}{4} a = d$

Combine $\frac{t}{4}$ and $a$ .

$- f^{3} + \frac{t a}{4} = d$

Add $f^{3}$ to both sides of the equation.

$\frac{t a}{4} = d + f^{3}$

Multiply both sides of the equation by $4$ .

$4 \frac{t a}{4} = 4 (d + f^{3})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{t a}{4} = 4 (d + f^{3})$

Rewrite the expression.

$t a = 4 (d + f^{3})$

Simplify the right side.

Apply the distributive property.

$t a = 4 d + 4 f^{3}$

Divide each term in $t a = 4 d + 4 f^{3}$ by $t$ and simplify.

Divide each term in $t a = 4 d + 4 f^{3}$ by $t$ .

$\frac{t a}{t} = \frac{4 d}{t} + \frac{4 f^{3}}{t}$

Simplify the left side.

Cancel the common factor of $t$ .

Cancel the common factor.

$\frac{t a}{t} = \frac{4 d}{t} + \frac{4 f^{3}}{t}$

Divide $a$ by $1$ .

$a = \frac{4 d}{t} + \frac{4 f^{3}}{t}$