Question

ch 9 algebraic expressions and the identities all formulas
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Answer 1

(a+b)2=a2+2ab+b2

(a−b)2=a2−2ab+b2

(a+b)(a–b)=a2–b2

(x+a)(x+b)=x2+(a+b)x+ab

(x+a)(x–b)=x2+(a–b)x–ab

(x–a)(x+b)=x2+(b–a)x–ab

(x–a)(x–b)=x2–(a+b)x+ab

(a+b)3=a3+b3+3ab(a+b)

(a–b)3=a3–b3–3ab(a–b)

(x+y+z)2=x2+y2+z2+2xy+2yz+2xz

(x+y–z)2=x2+y2+z2+2xy–2yz–2xz

(x–y+z)2=x2+y2+z2–2xy–2yz+2xz

(x–y–z)2=x2+y2+z2–2xy+2yz–2xz

x3+y3+z3–3xyz=(x+y+z)(x2+y2+z2–xy–yz−xz)

x2+y2=12[(x+y)2+(x–y)2]

(x+a)(x+b)(x+c)=x3+(a+b+c)x2+(ab+bc+ca)x+abc

x3+y3=(x+y)(x2–xy+y2)

x3–y3=(x–y)(x2+xy+y2)

x2+y2+z2−xy–yz–zx=12[(x−y)2+(y−z)2+(z−x)2]

Here it is pal.

Answer 2

$\begin{gathered}\boxed{\begin{array}{l}\boxed{\bigstar\:\:\textbf{\textsf{Algebric\:Identity}}\:\bigstar}\\\\\frak{1.}\bf\:(A+B)^{2} = A^{2} + 2AB + B^{2}\\\\\frak{2.}\sf\: (A-B)^{2} = A^{2} - 2AB + B^{2}\\\\\frak{3.}\bf\: A^{2} - B^{2} = (A+B)(A-B)\\\\\frak{4.}\sf\: (A+B)^{2} = (A-B)^{2} + 4AB\\\\\frak{5.}\bf\: (A-B)^{2} = (A+B)^{2} - 4AB\\\\\frak{6.}\sf\: (A+B)^{3} = A^{3} + 3AB(A+B) + B^{3}\\\\\frak{7.}\bf\:(A-B)^{3} = A^{3} - 3AB(A-B) + B^{3}\\\\\frak{8.}\sf\: A^{3} + B^{3} = (A+B)(A^{2} - AB + B^{2})\\\\\end{array}}\end{gathered}$