Question

(a + b) (az-ab+2
$(a + b)(a 2 - ab + {b}^{2}$
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Answer 1

Answer:

Step-by-step explanation:

Here is a list of Algebraic formulas –

a2 – b2 = (a – b)(a + b)

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

a2 + b2 = (a + b)2 – 2ab

(a – b)2 = a2 – 2ab + b2

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(a – b – c)2 = a2 + b2 + c2 – 2ab + 2bc – 2ca

(a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab(a – b)

a3 – b3 = (a – b)(a2 + ab + b2)

a3 + b3 = (a + b)(a2 – ab + b2)

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4

a4 – b4 = (a – b)(a + b)(a2 + b2)

a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

If n is a natural number an – bn = (a – b)(an-1 + an-2b+…+ bn-2a + bn-1)

If n is even (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)

If n is odd (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +an-3b2…- bn-2a + bn-1)

(a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….)