Question

if a^3 = b^3 + c^3 + d^3 , then least value of a is ?

Answer 1

PLS HELP

Answer 2

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

Answer 3

AALGEBRA

Answer 4

PLS HELP ME

Answer 5

HELP ME PLEASE

Answer 6

a<0 , b<0 , c<0

Answer 7

HOW

Answer 8

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 \(a\).

Answer 9

OK THANKS VERY MUCH

Answer 10

ANY1 HAVE A DIFFERENT ANSWER????????????///

Answer 11

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

Answer 12

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

Answer 13

no it is the full given

Answer 14

help me any1 else

Answer 15

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

Answer 16

can we do it with hit and trial method

Answer 17

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

Answer 18

i think we should refer options first

Answer 19

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

Answer 20

hey the answer is 9 i think

Answer 21

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

Answer 22

WHAT DO U ALL SAY ABOUT THIS ANSWER I THINK I M CORRECT

Answer 23

PLS SUGGEST

Answer 24

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

Answer 25

pls reply me

Answer 26

i just did hit and trial method

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

yes thats it its 3

Answer 39

YEAH THAT MAY BE

Answer 40

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

Answer 41

wait

Answer 42

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

Answer 43

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

Answer 44

but it is showing 6 as the answer

Answer 45

do b,c,d have to be postiive?

Answer 46

if not then 6 is wrong

Answer 47

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

Answer 48

ok, wc and thnx