Question

State true or false
$(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-5b)$=$19ca-37ab-19a^2$

• A. True
• B. False

Answer 1

The correct option is B False

We solve the given expression $(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-5b)$

$(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-5b)=\left[\right(a-b\left)\right(a-2b+5c\left)\right]-\left[\right(2a+3c\left)\right(a-b-c\left)\right]+\left[\right(6a+b\left)\right(2c-3a-5b\left)\right]=\left[a\right(a-2b+5c)-b(a-2b+5c\left)\right]-\left[2a\right(a-b-c)+3c(a-b-c\left)\right]+\left[6a\right(2c-3a-5b)+b(2c-3a-5b\left)\right]=(a^2-2ab+5ac-ab+2b^2-5bc)-(2a^2-2ab-2ac+3ac-3bc-3c^2)+(12ac-18a^2-30ab+2bc-3ab-5b^2)=(a^2+2b^2-3ab+5ac-5bc)-(2a^2-3c^2-2ab+ac-3bc)+(-18a^2-5b^2+12ac-33ab+2bc)=-19a^2-3b^2-3c^2-38ab+18ac-6bc$

Therefore, $(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-5b)=-19a^2-3b^2-3c^2-38ab+18ac-6bc$

Hence, $(a-2b+5c)(a-b)-(a-b-c)(2a+3c)+(6a+b)(2c-3a-5b)=19ca-37ab-19a^2$ is false.