## (y-2z)the whole square with explanation​

Question

(y-2z)the whole square with explanation​

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7 days 2021-09-11T13:48:01+00:00 2 Answers 0 views 0

1. ## CorrectQuestion: We have to evaluate it.

Use (a – b)^2 identity to solve it.

Since (a – b)^2 = a^2 + 2ab + b^2 Hence, y^2 + 4yz + 4z^2 is the solution.

## SomealgebraicIdentities:

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)

2. Some algebraic Identities:

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)