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

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

Mathematics
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PLS HELP
hi kushashwa,,Are you in algebra class or calculus?
AALGEBRA

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Other answers:

PLS HELP ME
HELP ME PLEASE
a<0 , b<0 , c<0
HOW
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 \(a\).
OK THANKS VERY MUCH
ANY1 HAVE A DIFFERENT ANSWER????????????///
WATCH MATH UR ANSWER IS NOT IN OPTION A)3 B)9 C)6 D) 12
Then you didn't put the problem completely. What do we know about a,b,c,d? we need more context.
no it is the full given
help me any1 else
The complete problem usually use a complete sentece like this: Let \(a,b,c,d\) be such and such .....
can we do it with hit and trial method
i thnk it is , if b=c=d then a^3 = b^3 + b^3 + b^3 =3b^3
i think we should refer options first
now if b=1, then a^3=3(1)^3 a^3=3
hey the answer is 9 i think
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
WHAT DO U ALL SAY ABOUT THIS ANSWER I THINK I M CORRECT
PLS SUGGEST
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
pls reply me
i just did hit and trial method
but where did you get those 8, 6 and 1?
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.
hence i found many pairs but i found 8, 6 and 1 as correct group . AM I CORRECT PLS TELL ME
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
M VERY SORRY TO ALL OF U THE ANSWER WILL BE 6 6^3 = 5^3 + 4^3 + 3^3 216 = 216
NO MARK IT CAN NOT BE THREE 3^3 IS NOT EQUAL TO ANY SUM OF THREE CUBES
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
TELL ME CLEARLY WHAT IS THE ANSWER. WHAT DO U THINK. I THINK 6 WOULD BE THE ANSWER
A)3 B)9 C)6 D) 12
say b=1,c=2 and d=3... also try b-0,c=1 and d=3
3 since 3^3=3^3+0^3+0^3
yes thats it its 3
YEAH THAT MAY BE
it is since there are no restrictions on a,b,c,d, right?
wait
i mean of the options for a that is the smallest but there can be smaller
like a->-infinity would be the smallest lol since a negative to an odd power is still negative and can be negative large
but it is showing 6 as the answer
do b,c,d have to be postiive?
if not then 6 is wrong
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
ok, wc and thnx

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