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

Question

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

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

Answers ( )

  1. Ava
    0
    2021-09-11T13:49:06+00:00

    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)

    0
    2021-09-11T13:49:40+00:00

    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)

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )