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sum of a and b is 1, find a^3 + b^3

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2 Answers

sum of a and b is 1, find a^3 + b^3
options:
1) 1 2)0 3) 3 4)none of these

a+b=1

(a+b)^3 = 1

Pascal’s Triangle
1
1 1
1 2 1
1 3 3 1 <== coefficients for binomial expansion

1^3 = (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = 1

a^3 + b^3 = 1 - 3a^2b - 3ab^2

a^3 + b^3 = 1 - 3ab(a + b) = 1-3ab(1)

a^3 + b^3 = 1-3ab = 1-3a(1-a) = 3a^2-3a+1

if 1 = 1-3ab, ab=0, so a or b is 0

So answer 1 works if ab=0.

if 0 = 3a^2-3a+1
(-3)^2-4(3)(1) < 0 ==> a imaginary

if 3 = 3a^2-3a+1
3a^2-3a-2=0
(-3)^2-4(3)(-2)=33
a=(3±√(33))/6
b=1-a=6/6-3/6±(-1)√(33)/6
b=(3±(-1)√(33))/6

a+b = ((3+√(33))/6)+((3–√(33))/6)=1/2+1/2=1 √

a^3+b^3 = ((3+√(33))/6)^3+((3–√(33))/6)^3

= (1/2)^3+3(1/2)^2(√(33))/6)+3(1/2)(√(33))/6)^2+(√(33))/6)^3
+(1/2)^3+3(1/2)^2(-√(33))/6)+3(1/2)(-√(33))/6)^2+(-√(33))/6)^3

= 2(1/2)^3+6(1/2)(√(33))/6)^2

= 1/4+3(33)/36 = 1/4+(11)/4 = 12/4 = 3 √

So answer 3 works if a = (3+√(33))/6 and b = (3–√(33))/6

So both answers 1 and 3 are correct.
Hi Channaraju,
 
a+b=1
 
Possibilities for a & b are,
 if a=1, b= 1-1=0  ,  or a=0, b=1 ,then a^3+b^3= 1^3 +0^3 = 1
 if a=2, b= 1-2=-1 , or a=-1 ,b=2 then   a^3+b^3 = 2^3+-1^3 = 8-1=7
 if a=3, b=-2,   or b=3,a=-2      then a^3+b^3 =3^3+-2^3=27-8=19......
 
so Ans: 1) 1 , a^3+b^3=1