Question

plz solve this
lesson = algebraic expressions and identities​

Answer 1

Calculation

(1) and (2) are alternative methods using substitution. You can go with expanding and grouping like terms.

(1) Let $t=x+y+z$.

Note that

• $t-2y-2z=x-y-z$
• $t-2z=x+y-z$
• $t-2y=x-y+z$

Then

$\implies 2x(t-2y-2z)+2y(t-2z)+2z(t-2y)$

Expand and group like terms for $t$.

$=t(2x+2y+2z)-4xy-4zx-4yz-4yz$

$=2t^2-4xy-8yz-4zx$

$=2(x^2+y^2+z^2+2xy+2yz+2zx)-4xy-8yz-4zx$

$=2x^2+2y^2+2z^2-4yz$

(2) Let $t=a+b+c$.

Note that

• $t-2b=a-b+c$
• $t-2c=a+b-c$
• $t-2b-2c=a-b-c$

$\implies 3a(t-2b)+2b(t-2c)-2c(t-2b-2c)$

$= 2a(t-2b)+2b(t-2c)-2c(t-2b-2c)+a(t-2b)$

Expand and group like terms for $t$.

$=t(2a+2b-2c)-4ab-4bc+4bc+4c^2+a(a-b+c)$

$=t(2a+2b-2c)-4ab+4c^2+a^2-ab+ca$

$=2(a+b+c)(a+b-c)-5ab+4c^2+a^2+ca$

$=2(a+b)^2-2c^2-5ab+4c^2+a^2+ca$

$=2(a^2+2ab+b^2)+2c^2-5ab+a^2+ca$

$=3a^2+2b^2+2c^2-ab+ca$

(3) F.O.I.L Method

$(a+b)(c-d)+(a-b)(c+d)+2(ac+bd)$

$=(ac-ad+bc-bd)+(ac+ad-bc-bd)+2(ac+bd)$

$=2ac-2bd+2ac+2bd$

$=4ac$

(4) Difference of two squares.

$(a+b+c)(a+b-c)-2ab$

$=(a+b)^2-c^2-2ab$

$=(a^2+2ab+b^2)-c^2-2ab$

$=a^2+b^2-c^2$