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· commented

· 1 Months ago

@heena Apply (a-b)³=a³-3a²b+3ab²-b³.

heena

first is the easiest but could you please explain how to put the two values in the questiion....

3 is the answer.

Method 01:

a^2 + 1 = a
a^2 = a - 1
a^12 = (a - 1)^6
a^6 = (a - 1)^3
a^12 + a^6 + 1 =3

Method 02:

a^2 + 1 = a
a^2 - a + 1 = 0
a = (1 +/- i sqrt 3)/2
Complex solutions:
0.5 + 0.86603 i
0.5 - 0.86603 i
a^12 + a^6 + 1 = 3

Method 03:

a^2 - a + 1 = 0
a = (-(-1) +/- sqrt((-1)^2 - 4(1)(1))) / (2*1)
a = (1 +/- sqrt(1 - 4)) / 2
a = (1 +/- sqrt(-3)) / 2
a = 0.5 +/- sqrt(-0.75) =3

Method 04:

a power 12 plus a power 6 plus 1 equal to (-1plus a) power 6 plus (-1 plus a) power 3 plus 1 ,........ Now, apply binomial theorem for that 2 brackets

solution please

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