Question

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

Answer 1

Correct Question:

$(y - 2z) {}^{2}$

We have to evaluate it.

Use (a – b)^2 identity to solve it.

Since (a – b)^2 = a^2 + 2ab + b^2

$= > (y - 2z) {}^{2} \\ \\ = y {}^{2} + 2 \times y \times 2z + ({2z}) {}^{2} \\ \\ = y {}^{2} + 4yz + 4z {}^{2}$

Hence, y^2 + 4yz + 4z^2 is the solution.

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)

Answer 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)