Solve 3[(4+3b)^2+2]+[2*(a-b)^2]=

$$3 [ ( 4 + 3 b ) ^ { 2 } + 2 ] + [ 2 \cdot ( a - b ) ^ { 2 } ] =$$

Evaluate

$$2\left(a-b\right)^{2}+3\left(3b+4\right)^{2}+6$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(4+3b\right)^{2}.

$$3\left(16+24b+9b^{2}+2\right)+2\left(a-b\right)^{2}$$

Add 16 and 2 to get 18.

$$3\left(18+24b+9b^{2}\right)+2\left(a-b\right)^{2}$$

Use the distributive property to multiply 3 by 18+24b+9b^{2}.

$$54+72b+27b^{2}+2\left(a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$54+72b+27b^{2}+2\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply 2 by a^{2}-2ab+b^{2}.

$$54+72b+27b^{2}+2a^{2}-4ab+2b^{2}$$

Combine 27b^{2} and 2b^{2} to get 29b^{2}.

$$54+72b+29b^{2}+2a^{2}-4ab$$

Expand

$$2a^{2}-4ab+29b^{2}+72b+54$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(4+3b\right)^{2}.

$$3\left(16+24b+9b^{2}+2\right)+2\left(a-b\right)^{2}$$

Add 16 and 2 to get 18.

$$3\left(18+24b+9b^{2}\right)+2\left(a-b\right)^{2}$$

Use the distributive property to multiply 3 by 18+24b+9b^{2}.

$$54+72b+27b^{2}+2\left(a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$54+72b+27b^{2}+2\left(a^{2}-2ab+b^{2}\right)$$

Use the distributive property to multiply 2 by a^{2}-2ab+b^{2}.

$$54+72b+27b^{2}+2a^{2}-4ab+2b^{2}$$

Combine 27b^{2} and 2b^{2} to get 29b^{2}.

$$54+72b+29b^{2}+2a^{2}-4ab$$