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Formula
Factorize the expression
Answer
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$x ^{ 3 } +8 =$
$\left ( x + 2 \right ) \left ( x ^ { 2 } - 2 x + 4 \right )$
Arrange the expression in the form of factorization..
$\color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 8 }$
$ $ Use the factoring formula, $ a^{3} + b^{3} = \left(a+b\right)\left(a^{2}-ab + b^{2}\right)$
$\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 4 } \right )$
Solution search results
search-thumbnail-Synthefic DivisPon 

$0=2x^{2}-x+5$ $x^{3}+6x^{2}-7x-8\div x+7$ 
$x+4$ $28^{3}$ $2$ remainder $8er=-20$ $-$ $2$ $-7y-8$ $x+3\sqrt{\times 3} +6\dfrac {x} {+}x^{2}$ $\dfrac {\times 3+} {-x^{2}}k-$ 
$3$ $2$ $8$ $\div 20$ $20$ $ac28$ $=x^{2}-x$ $\dfrac {-f_{x}-8} {-8}$ 
$remaixkr=-8$ 
$4x^{3}+5x^{2}-6\times +8:x-3$ $③$ $3x^{4+4x^{2}-2x-11\dfrac {2} {2}x-2}$ 
$2x^{3}+8x^{2}-3\div x+5$ $②$ ☺ $x^{4}-3x^{3}+5x^{2}+2\div x-3$ 
$E$ $x^{3}+8x^{2}-6x-112x+7$
10th-13th grade
Calculus
search-thumbnail-Algebra 
12. For the following functions, verify that f(x) and $f^{-1}\left(x\right)$ are indeed inverse functions. That is, 
$1\left(x\right)$ 
show that $\left(f0f^{-1}\right)\left(x\right)=x$ and $\left(f^{-1}0f\right)\left(x\right)=x$ 
$f\left(x\right)=\sqrt [3] {x+8} ,f^{-1}\left(x\right)=x^{3}-8$ 
Choose the correct answer below. 
$○A.$ $\left(fof^{-1}\right)\left(x\right)=\sqrt [3] {\left(x^{3}-8\right)+8} =\sqrt [3] {x^{3}} =x$ 
BO . B. $\left(fof^{-1}\right)\left(x\right)=\left(\sqrt [3] {x-8} \right)^{3}+8=x-8+8=x$ 
OC. $\left({f_{0f}}^{-1}\right)\left(x\right)=\sqrt [3] {\left(x^{3}+8\right)-8} =\sqrt [3] {x^{3}} =x$ 
$○D$ D. $\left(fof^{-1}\right)\left(x\right)=\left(\sqrt [3] {x+8} \right)^{3}-8=x+8-8=x$ 
Choose the correct answer below. 
$○A$ A. $\left(f^{-1}°f\right)\left(x\right)=\left(\sqrt [3] {x+8} \right)^{3}-8=x+8-8=x$ 
O B. $\left(f^{-1}°f\right)\left(x\right)=\sqrt [3] {\left(x^{3}+8\right)-8} =3\sqrt{x^{3}} =x$ 
$C,$ $\left(f^{-1}°f\right)\left(x\right)=\sqrt [3] {\left(x^{3}-8\right)+8} =\sqrt [3] {x^{3}} =x$ 
$D.$ $\left(f^{-1}°f\right)\left(x\right)=\left(\sqrt [3] {x-8} \right)^{3}+8=x-8+8=x$
1st-6th grade
Algebra
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