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

- anonymous

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

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- anonymous

PLS HELP

- anonymous

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

- anonymous

AALGEBRA

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## More answers

- anonymous

PLS HELP ME

- anonymous

HELP ME PLEASE

- anonymous

a<0 , b<0 , c<0

- anonymous

HOW

- 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 \(a\).

- anonymous

OK THANKS VERY MUCH

- anonymous

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

- anonymous

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

- watchmath

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

- anonymous

no it is the full given

- anonymous

help me any1 else

- watchmath

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

- anonymous

can we do it with hit and trial method

- anonymous

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

- anonymous

i think we should refer options first

- anonymous

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

- anonymous

hey the answer is 9 i think

- 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

- anonymous

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

- anonymous

PLS SUGGEST

- 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

- anonymous

pls reply me

- anonymous

i just did hit and trial method

- anonymous

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

- 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.

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- 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

- anonymous

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

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

yes thats it its 3

- anonymous

YEAH THAT MAY BE

- myininaya

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

- anonymous

wait

- myininaya

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

- myininaya

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

- anonymous

but it is showing 6 as the answer

- myininaya

do b,c,d have to be postiive?

- myininaya

if not then 6 is wrong

- 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

- anonymous

ok, wc and thnx

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