CelsiusSolving

\((a-b)^3=a^3-3a^2b+3ab^2-b^3\)=

\(1-3ab(a-b)\)=

\(1-3ab\)=1

3ab=0,

ab=0

Because ab=0, we know \((a-b)^2\)=\(a^2+b^2+0\)=1

Because of this, \((a+b)^2=1+0=1\), so (a+b) is +/-1

Finally because a+b is 1, and one of either a or b is 0, because ab is 0, the possible pairings of (a,b) are

(0,1), or (1,0)

Hope this helps!

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\(9^{x}-2\cdot 3^{x}+1\):

Set a = \(3^x\)

\(a^2-2a+1=0\)

\((a-1)^2=0 \)

a=1, so \(3^x\)=1

That means x=0 because anything to the 0th power is one

Hope this helps! 

 🤙 🤙 🤙

N mod 10 = a

N mod 2 = b

N mod 20 =?

2 and 10 are not coprime. 10 is multiple of 2.

If a is odd then b=1.

If a is even then b=0.

We don’t need b.

N mod 20 ={a, a+10}