(a+b)2 + (a-b) 2 prove that
Answers
Step-by-step explanation:
a^2+b^2+2ab+a^2+b^2-2ab
a^2+b^2+a^2+b^2
2a^2+2b^2
2(a^2+b^2)
Answer:
How do I prove that (a+b)² × (a-b)² = (a²-b²)² ?
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[math](a+b)^2×(a-b)^2=(a^2-b^2)2[/math]
Write the formulas of [math](a+b)^2[/math] and [math](a-b)^2[/math]
i.e.,
[math](a+b)^2=a^2+2ab+b^2[/math]
[math](a-b)^2=a^2-2ab+b^2[/math]
Now, put the formulas,
[math](a+b)^2×(a-c)^2\;\;\implies(a^2+2ab+b^2)(a^2-2ab+b^2)[/math]
Multiply each term of first factor by each term of second factor,
[math]a^4-2a^3b+a^2b^2+2a^3b-4a^2b^2+2ab^3+b^2a^2-2ab^3+b^4[/math]
Cancel out the like terms with opposite signs,
[math]a^4-\not 2\not a^3\not b+a^2b^2+\not 2\not a^3\not b-4a^2b^2+\not 2\not a\not b^3+b^2a^2-\not 2\not a\not b^3+b^4[/math]
[math]a^4+a^2b^2-4a^2b^2+b^2a^2+b^4[/math]
Rewrite it as,
([math]a^2)^2-2a^2b^2+(b^2)^2\;\;\;\;\;\;\;\;\;\;\;(\because a^2b^2-4a^2b^2+b^2a^2=-2a^2b^2)[/math]
Now the expression is in the form of [math](a-b)^2[/math]
where, [math]a=a^2[/math] and [math]b=b^2 [/math]
Therefore,
([math]a^2)^2-2a^2b^2+(b^2)^2=(a^2-b^2)2[/math]