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Question Number 215092 by BaliramKumar last updated on 28/Dec/24

Commented by BaliramKumar last updated on 28/Dec/24

I want to find the value of a and b

$${I}\:{want}\:{to}\:{find}\:{the}\:{value}\:{of}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}} \\ $$

Commented by BaliramKumar last updated on 28/Dec/24

a^3 −b^3 = 125 a^2 −b^2 = 25 a+b = 5 a^2 + ab + b^(2 ) = ?

$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} \:=\:\mathrm{125} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$${a}+{b}\:=\:\mathrm{5} \\ $$$${a}^{\mathrm{2}} \:+\:{ab}\:+\:{b}^{\mathrm{2}\:} \:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 28/Dec/24

a^2 −b^2 =25⇒(a−b)(a+b)=25 (a−b)(5)=25⇒a−b=5 { ((a+b=5)),((a−b=5)) :}⇒a=5,b=0 a^2 +ab+b^2 =5^2 +5×0+0^2 =25

$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{25}\Rightarrow\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{25} \\ $$$$\left({a}−{b}\right)\left(\mathrm{5}\right)=\mathrm{25}\Rightarrow{a}−{b}=\mathrm{5} \\ $$$$\begin{cases}{{a}+{b}=\mathrm{5}}\\{{a}−{b}=\mathrm{5}}\end{cases}\Rightarrow{a}=\mathrm{5},{b}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +\mathrm{5}×\mathrm{0}+\mathrm{0}^{\mathrm{2}} =\mathrm{25} \\ $$

Commented by A5T last updated on 28/Dec/24

We can also do this,so we get a contradiction. a^3 −b^3 =(a−b)(a^2 +ab+b^2 )=625 a^2 −b^2 =(a−b)(a+b)=25 a+b=5⇒a−b=((25)/5)=5 ⇒a^2 +ab+b^2 =((625)/5)=125

$${We}\:{can}\:{also}\:{do}\:{this},{so}\:{we}\:{get}\:{a}\:{contradiction}. \\ $$$${a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)=\mathrm{625} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({a}−{b}\right)\left({a}+{b}\right)=\mathrm{25} \\ $$$${a}+{b}=\mathrm{5}\Rightarrow{a}−{b}=\frac{\mathrm{25}}{\mathrm{5}}=\mathrm{5} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} =\frac{\mathrm{625}}{\mathrm{5}}=\mathrm{125} \\ $$

Commented by Frix last updated on 28/Dec/24

We′ve got 3 equations for 2 unknowns and all must be satisfied. (1) b=5−a (2) a^2 =25+b^2 a^2 =25+(5−a)^2 a^2 =50−10a+a^2 ⇒ a=5 ⇒ b=0 (3) Test a^3 −b^3 =625 ⇒ wrong ⇒ No solution

$$\mathrm{We}'\mathrm{ve}\:\mathrm{got}\:\mathrm{3}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{2}\:\mathrm{unknowns}\:\mathrm{and} \\ $$$$\mathrm{all}\:{must}\:\mathrm{be}\:\mathrm{satisfied}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:{b}=\mathrm{5}−{a} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{a}^{\mathrm{2}} =\mathrm{25}+{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} =\mathrm{25}+\left(\mathrm{5}−{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} =\mathrm{50}−\mathrm{10}{a}+{a}^{\mathrm{2}} \:\Rightarrow\:{a}=\mathrm{5}\:\Rightarrow\:{b}=\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\mathrm{Test}\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\mathrm{625}\:\Rightarrow\:\mathrm{wrong} \\ $$$$\Rightarrow \\ $$$$\mathrm{No}\:\mathrm{solution} \\ $$

Commented by BaliramKumar last updated on 29/Dec/24

i think question is wrong

$${i}\:{think}\:{question}\:{is}\:{wrong} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/24

I think that the question has come before this time also.

$${I}\:{think}\:{that}\:{the}\:{question}\:{has}\:{come} \\ $$$$\:{before}\:{this}\:{time}\:{also}. \\ $$

Commented by A5T last updated on 29/Dec/24

Q211393

$${Q}\mathrm{211393} \\ $$

Commented by Frix last updated on 29/Dec/24

A question can not be wrong. The proposed answers are wrong.

$$\mathrm{A}\:\mathrm{question}\:\mathrm{can}\:\mathrm{not}\:\mathrm{be}\:\mathrm{wrong}. \\ $$$$\mathrm{The}\:\mathrm{proposed}\:\mathrm{answers}\:\mathrm{are}\:\mathrm{wrong}. \\ $$

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