B⁶=3⁶ B⁶-3⁶=0 (B³)²-(3³)²=0 (B³+3³)(B³-3³)=0. [A²-B²=(A+B)(A-b)] Using a³+b³= (a+b)(a²-ab+b²) And a³-b³= (a-b)(a²+ab+b²), we get (B+3)(B-3)(B²-3B+9)(B²+3B+9)=0 So we get B+3=0, B-3= 0 for B= {+3,-3} And B²-3b+9=0 a=1, b=-3, c=9 B = [-b+/-√(b²-4ac)]/2 B=[-(-3)+/-√((-3)²-4×9)]/2 B= [3+/-√(9-36)]/2 B=[3+/-√(-27)]/2 Since we can't calculate negative square roots, we have to use the complex number i=√-1 √-27= √(27×-1) Using the property √(a×b)= √a√b, we get √-27=√27 ×√-1 √-27=√27i So we get two more solutions for B, extending the solution set to= {3, -3, (3+√27i)/2, (3-√27i)/2} Now solving B²+3B+9=0 a= 1, b=3, c=9 B= [-b+/-√(b²-4ac)]/2 B= [-3+/-√27i] Making the final solution set= {3, -3, (3+√27i)/2, (3-√27i)/2, (-3+√27i), (-3-√27i)} We could have taken √27 to be 5.2 and simplified more for a simplified but approximate answer, but I let it stay as it is because its depend on the type of question we are solving, say a real life problem will only have the solution +3 and -3 but in a competitive exam, we may require the complex roots. Asad, if you see this, pls pin me
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B⁶=3⁶
B⁶-3⁶=0
(B³)²-(3³)²=0
(B³+3³)(B³-3³)=0. [A²-B²=(A+B)(A-b)]
Using a³+b³= (a+b)(a²-ab+b²)
And a³-b³= (a-b)(a²+ab+b²), we get
(B+3)(B-3)(B²-3B+9)(B²+3B+9)=0
So we get B+3=0, B-3= 0 for B= {+3,-3}
And
B²-3b+9=0
a=1, b=-3, c=9
B = [-b+/-√(b²-4ac)]/2
B=[-(-3)+/-√((-3)²-4×9)]/2
B= [3+/-√(9-36)]/2
B=[3+/-√(-27)]/2
Since we can't calculate negative square roots, we have to use the complex number i=√-1
√-27= √(27×-1)
Using the property √(a×b)= √a√b, we get
√-27=√27 ×√-1
√-27=√27i
So we get two more solutions for B, extending the solution set to=
{3, -3, (3+√27i)/2, (3-√27i)/2}
Now solving B²+3B+9=0
a= 1, b=3, c=9
B= [-b+/-√(b²-4ac)]/2
B= [-3+/-√27i]
Making the final solution set= {3, -3, (3+√27i)/2, (3-√27i)/2, (-3+√27i), (-3-√27i)}
We could have taken √27 to be 5.2 and simplified more for a simplified but approximate answer, but I let it stay as it is because its depend on the type of question we are solving, say a real life problem will only have the solution +3 and -3 but in a competitive exam, we may require the complex roots.
Asad, if you see this, pls pin me
It's really easy to overlook the complex solutions by simply inferring b = 3 or b = +- 3. Beautiful exercise!
Totally agree!
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b = 3 😂
b=3
why not -3?
😂! the solution can be seen without calculation! b=3, b=0, b=-3 😂😂😂
There are 6 solutions to this
ok dear
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yes dear
b=3