$1$ . Let $a, b>0$ satisfy $a^{3}+b^{3}=a-b$ , then $\newline$ (a) $a^{2}+b^{2}=1$ $\newline$ (b) $a^{2}+a b+b^{2}<1$ $\newline$ (c) $a^{2}+b^{2}>1$ $\newline$ (d) none of these

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Algebra 2

Evaluate expression when a complex numbers and a variable term is given

Full solution

1

. Let

a, b>0

satisfy

a^{3}+b^{3}=a-b

, then

\newline

(a)

a^{2}+b^{2}=1

\newline

(b)

a^{2}+a b+b^{2}<1

\newline

(c)

a^{2}+b^{2}>1

\newline

(d) none of these

Calculate using identity: Calculate $a^3 + b^3$ using the identity $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ . $\newline$ - Given: $a^3 + b^3 = a - b$ $\newline$ - Identity: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ $\newline$ - Set equal: $(a + b)(a^2 - ab + b^2) = a - b$
Simplify the equation: Simplify the equation assuming $a + b \neq 0$ (since a, b > $0$ ). $\newline$ - Divide both sides by $a + b$ : $a^2 - ab + b^2 = \frac{a - b}{a + b}$
Check option (a): Check each option by substituting and simplifying. $\newline$ (a) $a^2 + b^2 = 1$ $\newline$ - $a^2 + b^2 = 1$ implies $a^2 - ab + b^2 = 1 - ab$ $\newline$ - Compare: $1 - ab = \frac{a - b}{a + b}$ $\newline$ - This equation is not generally true without specific values for a and b.
Check option (b): Check option (b). $\newline$ (b) $a^2 + ab + b^2 < 1$ $\newline$ - $a^2 + ab + b^2 = (a^2 - ab + b^2) + 2ab$ $\newline$ - Substitute: $\frac{a - b}{a + b} + 2ab < 1$ $\newline$ - This inequality depends on the values of a and b, not generally provable without additional information.
Check option (c): Check option (c). $\newline$ (c) $a^2 + b^2 > 1$ $\newline$ - $a^2 + b^2 = (a^2 - ab + b^2) + ab$ $\newline$ - Substitute: $\frac{a - b}{a + b} + ab > 1$ $\newline$ - This inequality also depends on specific values of a and b, not generally provable.
Conclude the results: Conclude that none of the options can be universally proven with the given information. $\newline$ - None of the options $(a)$ , $(b)$ , or $(c)$ can be definitively proven true for all positive $a$ and $b$ satisfying the original equation.