Question

(a+b)²= a²+2ab+b². a²-b² = (a+b)(a-b)​

Answer 1

Answer:

Identity I: (a + b^)2 = a^2 + 2ab + b^2

Identity II: (a – b)^2 = a^2 – 2ab + b^2

Identity III: a^2 – b^2= (a + b)(a – b)

Identity IV: (x + a)(x + b) = x^2 + (a + b) x + ab

Identity V: (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

Identity VI: (a + b)^3 = a^3 + b^3 + 3ab (a + b)

Identity VII: (a – b)^3 = a^3 – b^3 – 3ab (a – b)

Identity VIII: a^3 + b^3 + c^3 – 3abc = (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)

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Answer 2

Answer:

The distributive property says for numbers A, B, C,

A × (B + C) = (A × B) + (A × C)

How does this apply to (a + b) × (a + b)?

a + b is just a number, so it is valid to treat (a + b) as A, B, or C in the distributive property, because the property *always* works when A, B, C are numbers.

So let's connect our expression to the distributive property, by setting a and b to terms with A, B, C.

A = a + b

B = a

C = b

After plugging in these values,

A × (B + C) = (A × B) + (A × C) becomes

(a + b) × (a + b) = ((a + b) × a) + ((a + b) × b)

Since the order of multiplication does not matter, we can rewrite the right side of that equation…

( a + b) × (a + b) = (a × (a + b)) + (b × (a + b))

Now, using the distributive property more simply,

(a + b) × (a + b) = (a×a + a×b) + (b×a + b×b)

The parentheses on the right side do not change order of operations, because the order that we add all of the multiplied terms up does not matter, so

(a + b) × (a + b) = a×a + a×b + b×a + b×b

Of course, b×a can be rewritten as a×b, so

(a + b) × (a + b) = a×a + a×b + a×b + b×b

Now we can combine the two like-terms of a×b

(a + b) × (a + b) = a×a + 2×a×b + b×b

Which is more common written as

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