how do you do this math problem (x+7)3 =x3 + 133?

Answer 1

It is pretty cool to see that it is a quadratic equation after all:

Use (x + a)^3 = x^3 + 3ax^2 + 3a^2*x + a^3 with a=7

x^3 + 21x^2 + 147x + 343 = x^3 + 133 // - x^3 + 133

21x^2 + 147x + 210 = 0 // Let's divide it by 21

x^2 + 7x + 10 = 0

Now, I can see without the whole procedrue of solving a quadratic equation that

(x + 2)(x + 5) = 0

x1 = -2
x2 = -5

Answer 2

do u mean (x+7)^3 =x^3 + 133?
if so then u will need to expand (x+7)^3 either with the aid of a pascal triangle or using binomial expansion

so (x+7)^3=x^3+3(x^2)*7+3x(7^2)+7^3=x^3+21x^2+147x+343

so we have x^3+21x^2+147x+343=x^3 + 133

so 21x^2+147x+210=0
x^2+7x+10=0
(x+5)(x+2)=o
x=-5 or x=-2

Answer 3

(x+7)^3 =x^3 + 133... three represents cube then this is the solution,

first, get (x+7)^2 which is (x^2+14x+49) multiply this by (x+7) to get (x+7)^3 equals to (x^3+21x^2+147x+343), equate this equation with x^3 + 133

x^3+21x^2+147x+343 = x^3 + 133

In general form, 21x^2+147x+210 = 0
divide this by 21, x^2+7x+10 = 0
by factorization, (x+5)(x+2)=0

x = -5 and -2

Answer 4

(x + 7)3 = x3 + 133
distribute
3x + 21 = 3x + 133
there is no solution
reductio ad absurdum

Answer 5

is it (x+7)* 3 ??

Answer 6

3(x+3)=3x+133
3x+9=3x+133
3x-3x=133-9
x dont exist

Answer 7

first of all you should know what is the value of x then i can answer your question!!