Solve (a-15)^2quad{(7-8)^3[(5*6)-(4*7)]}

$$( a - 15 ) ^ { 2 } \quad \{ ( 7 - 8 ) ^ { 3 } [ ( 5 \times 6 ) - ( 4 \times 7 ) ] \}$$

Evaluate

$$-2\left(a-15\right)^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-15\right)^{2}.

$$\left(a^{2}-30a+225\right)\left(7-8\right)^{3}\left(5\times 6-4\times 7\right)$$

Subtract 8 from 7 to get -1.

$$\left(a^{2}-30a+225\right)\left(-1\right)^{3}\left(5\times 6-4\times 7\right)$$

Calculate -1 to the power of 3 and get -1.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(5\times 6-4\times 7\right)$$

Multiply 5 and 6 to get 30.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(30-4\times 7\right)$$

Multiply 4 and 7 to get 28.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(30-28\right)$$

Subtract 28 from 30 to get 2.

$$\left(a^{2}-30a+225\right)\left(-1\right)\times 2$$

Multiply -1 and 2 to get -2.

$$\left(a^{2}-30a+225\right)\left(-2\right)$$

Use the distributive property to multiply a^{2}-30a+225 by -2.

$$-2a^{2}+60a-450$$

Expand

$$-2a^{2}+60a-450$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-15\right)^{2}.

$$\left(a^{2}-30a+225\right)\left(7-8\right)^{3}\left(5\times 6-4\times 7\right)$$

Subtract 8 from 7 to get -1.

$$\left(a^{2}-30a+225\right)\left(-1\right)^{3}\left(5\times 6-4\times 7\right)$$

Calculate -1 to the power of 3 and get -1.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(5\times 6-4\times 7\right)$$

Multiply 5 and 6 to get 30.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(30-4\times 7\right)$$

Multiply 4 and 7 to get 28.

$$\left(a^{2}-30a+225\right)\left(-1\right)\left(30-28\right)$$

Subtract 28 from 30 to get 2.

$$\left(a^{2}-30a+225\right)\left(-1\right)\times 2$$

Multiply -1 and 2 to get -2.

$$\left(a^{2}-30a+225\right)\left(-2\right)$$

Use the distributive property to multiply a^{2}-30a+225 by -2.

$$-2a^{2}+60a-450$$