## anonymous 5 years ago if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

1. anonymous

PLS HELP

2. anonymous

hi kushashwa,,Are you in algebra class or calculus?

3. anonymous

AALGEBRA

4. anonymous

PLS HELP ME

5. anonymous

6. anonymous

a<0 , b<0 , c<0

7. anonymous

HOW

8. watchmath

Assuming that a,b,c,d are all positive then we have $$a^3\geq 3\sqrt[3]{b^3c^3d^3}=3bcd$$ Hence $a\geq \sqrt[3]{3bcd}$ So $$\sqrt[3]{3bcd}$$ is the minimum value of $$a$$.

9. anonymous

OK THANKS VERY MUCH

10. anonymous

11. anonymous

WATCH MATH UR ANSWER IS NOT IN OPTION A)3 B)9 C)6 D) 12

12. watchmath

Then you didn't put the problem completely. What do we know about a,b,c,d? we need more context.

13. anonymous

no it is the full given

14. anonymous

help me any1 else

15. watchmath

The complete problem usually use a complete sentece like this: Let $$a,b,c,d$$ be such and such .....

16. anonymous

can we do it with hit and trial method

17. anonymous

i thnk it is , if b=c=d then a^3 = b^3 + b^3 + b^3 =3b^3

18. anonymous

i think we should refer options first

19. anonymous

now if b=1, then a^3=3(1)^3 a^3=3

20. anonymous

hey the answer is 9 i think

21. anonymous

because by hit and trial method 9 is the least number that satisfies the given condition 9^3 = 8^3 + 6^3 + 1^3 729 = 512 + 216 + 1 729 = 729

22. anonymous

23. anonymous

PLS SUGGEST

24. anonymous

because by hit and trial method 9 is the least number that satisfies the given condition 9^3 = 8^3 + 6^3 + 1^3 729 = 512 + 216 + 1 729 = 729 IS THIS CORRECT SOLUTION

25. anonymous

26. anonymous

i just did hit and trial method

27. anonymous

but where did you get those 8, 6 and 1?

28. anonymous

if a^3 = b^3 + c^3 + d^3 hence i thought that a must be greater than b, c and d. so if a = 9 hence b , c and d will be among 1, 2, 3, 4 , 5, 6, 7, 8 i took the cubes of these numbers and took the numbers whose sum contains unit digit as 9.

29. anonymous

hence i found many pairs but i found 8, 6 and 1 as correct group . AM I CORRECT PLS TELL ME

30. anonymous

i think it is 3,,, try b=1, c=0 d=0, then try c=1,b=0,d=0, then try d=1,b=0,c=0

31. anonymous

M VERY SORRY TO ALL OF U THE ANSWER WILL BE 6 6^3 = 5^3 + 4^3 + 3^3 216 = 216

32. anonymous

NO MARK IT CAN NOT BE THREE 3^3 IS NOT EQUAL TO ANY SUM OF THREE CUBES

33. anonymous

lol...well thats the least i can see ...lol,, so accordingly b cant be equal to c cant be equal to d,....ok....we need to complete the prob specifically...lol

34. anonymous

TELL ME CLEARLY WHAT IS THE ANSWER. WHAT DO U THINK. I THINK 6 WOULD BE THE ANSWER

35. anonymous

A)3 B)9 C)6 D) 12

36. anonymous

say b=1,c=2 and d=3... also try b-0,c=1 and d=3

37. myininaya

3 since 3^3=3^3+0^3+0^3

38. anonymous

yes thats it its 3

39. anonymous

YEAH THAT MAY BE

40. myininaya

it is since there are no restrictions on a,b,c,d, right?

41. anonymous

wait

42. myininaya

i mean of the options for a that is the smallest but there can be smaller

43. myininaya

like a->-infinity would be the smallest lol since a negative to an odd power is still negative and can be negative large

44. anonymous

but it is showing 6 as the answer

45. myininaya

do b,c,d have to be postiive?

46. myininaya

if not then 6 is wrong

47. anonymous

i also don't know the question that was given i typed i think the question is wrong it should mention the whole context thanks very much

48. anonymous

ok, wc and thnx