Solve 12+3a+3(a)^(2)+3a-3a+3b-3(a)^(2)-36+23+3(b)^(3) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$12+3a+3 { a }^{ 2 } +3a-3a+3b-3 { a }^{ 2 } -36+23+3 { b }^{ 3 }$$

Evaluate

$3a+3b^{3}+3b-1$

Solution Steps

Combine $3a$ and $3a$ to get $6a$.

$$12+6a+3a^{2}-3a+3b-3a^{2}-36+23+3b^{3}$$

Combine $6a$ and $-3a$ to get $3a$.

$$12+3a+3a^{2}+3b-3a^{2}-36+23+3b^{3}$$

Combine $3a^{2}$ and $-3a^{2}$ to get $0$.

$$12+3a+3b-36+23+3b^{3}$$

Subtract $36$ from $12$ to get $-24$.

$$-24+3a+3b+23+3b^{3}$$

Add $-24$ and $23$ to get $-1$.

$$-1+3a+3b+3b^{3}$$

Differentiate w.r.t. a

$3$