Question

what is the formula of a + b + c + d + whole square​

Answer 1

$\displaystyle \sf {(a + b + c + d)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}+ 2ab + 2ac + 2ad + 2bc + 2bd+ 2cd$

Given :

$\displaystyle \sf {(a + b + c + d)}^{2}$

To find :

The formula

Solution :

Step 1 of 2 :

Write down the expression

Here the given expression is

$\displaystyle \sf {(a + b + c + d)}^{2}$

Step 2 of 2 :

Find the formula

We are aware of the identity that

$\displaystyle \sf {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

Putting x = a + b and y = c + d we get

$\displaystyle \sf {(a + b + c + d)}^{2}$

$\displaystyle \sf = {(a + b)}^{2} + {(c + d)}^{2} + 2(a + b)(c + d)$

$\displaystyle \sf = {a}^{2} + 2ab + {b}^{2} + {c}^{2} + 2cd + {d}^{2} + 2ac + 2ad + 2bc + 2bd$

$\displaystyle \sf = {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2}+ 2ab + 2ac + 2ad + 2bc + 2bd+ 2cd$

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Answer 2

Answer:

$(a + b + c + d)^2$ $= a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)$

Step-by-step explanation:

The formula for $(a + b + c + d)^2$ is$(a + b + c + d)$ multiplied by itself, also known as squaring the expression. This can be written as:

$(a + b + c + d)^2$$= a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)$

When you multiply an integer (a “whole” number, positive, negative or zero) times itself, the resulting product is called a square number, or a perfect square or simply “a square.” So, 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, and so on, are all square numbers.

More formally: A square number is a number of the form n × n or n2 where n is any integer.

when we talk about let's say $(a+b)^2$,itwill expanded by a's square ,b's square plus two times of the their product with each other :

$a^2+b^2+2ab$.likewise  in the above case ,we will individually do the squares of every variable and add them with their multiplications/products.

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