Cube

Algebra Level 2

Find the value of $x$ .

$x =$ $99^{3} + 3(99)^{2} + 3(99) + 1$ .

2 solutions

A Former Brilliant Member
Sep 20, 2016

$\implies x =99^{3} + 3(99)^{2} + 3(99) + 1$

$x=(99+1)^3=100^3=\boxed{1000000}$ .

let y=99

so:

x=99^3+(3)(99)^2+(3)(99)+1 x=y^3+3y^2+3y+1 x=(y+1)^3 x=[(99)+1]^3 x=(100)^3 x=1000000

stephen liado - 4 years, 8 months ago

. .
Feb 19, 2021

$x = 99 ^ { 3 } + 3 \times 99 ^ { 2 } + 3 \times 99 + 1 \rightarrow 970299 + 29403 + 297 + 1 \Rightarrow \boxed { 1000000 } .$