anonymous
  • anonymous
if a^3 = b^3 + c^3 + d^3 , then least value of a is ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
PLS HELP
anonymous
  • anonymous
hi kushashwa,,Are you in algebra class or calculus?
anonymous
  • anonymous
AALGEBRA

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

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