(1) { ((x^3 +y^6 = 91)),((x+y^2 = 7 )) :} find x−y^6 . (2) 2a+(2/a) = 8 ⇒ ((a^6 +1)/a^3 ) ?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 104037 by bramlex last updated on 19/Jul/20
$(1) { ((x^3 +y^6 = 91)),((x+y^2 = 7 )) :} find x−y^6 . (2) 2a+(2/a) = 8 ⇒ ((a^6 +1)/a^3 ) ?$
$(1) {\begin{cases} x^{3} + y^{6} = 91 \\ x + y^{2} = 7 \end{cases}$ $f i n d x - y^{6} .$ $(2) 2 a + \frac{2}{a} = 8 \Rightarrow \frac{a^{6} + 1}{a^{3}} ?$
	Answered by bemath last updated on 19/Jul/20

	$(2) a + \frac{1}{a} = 4 \Rightarrow \frac{a^{2} + 1}{a} = 4$ $a^{2} + 1 = 4 a \Rightarrow {(a^{2} + 1)}^{3} = 64 a^{3}$ $\Rightarrow a^{6} + 3 a^{4} + 3 a^{2} + 1 = 64 a^{3}$ $a^{6} + 1 = 64 a^{3} - 3 a^{4} - 3 a^{2}$ $t h e n \frac{a^{6} + 1}{a^{3}} = \frac{64 a^{3} - 3 a^{2} (a^{2} + 1)}{a^{3}}$ $= \frac{64 a - 3 (a^{2} + 1)}{a} = 64 - 3 (\frac{4 a}{a})$ $= 64 - 12 = 52 ◼$
	Answered by bramlex last updated on 19/Jul/20
	$(1) (x+y^2 )^3 = 7^3 ⇒x^3 +3x^2 y^2 +3xy^4 +y^6 = 343 ⇒91+3x^2 y^2 +3xy^4 = 343 3xy^2 (x+y^2 ) = 252 ⇒3xy^2 =((252)/7) xy^2 = 12 ⇒y^2 = ((12)/x) substitute to x+y^2 =7 ⇒x+((12)/x) = 7 ⇒x^2 −7x+12=0 { ((x=3 ∧y^2 =4)),((x=4∧y^2 =3)) :} then x−y^6 = 3−4^3 or 4−3^3$
	$(1) {(x + y^{2})}^{3} = 7^{3} \Rightarrow x^{3} + 3 x^{2} y^{2} + 3 {x y}^{4} + y^{6}$ $= 343$ $\Rightarrow 91 + 3 x^{2} y^{2} + 3 {x y}^{4} = 343$ $3 {x y}^{2} (x + y^{2}) = 252 \Rightarrow 3 {x y}^{2} = \frac{252}{7}$ ${x y}^{2} = 12 \Rightarrow y^{2} = \frac{12}{x}$ $s u b s t i t u t e t o x + y^{2} = 7$ $\Rightarrow x + \frac{12}{x} = 7 \Rightarrow x^{2} - 7 x + 12 = 0$ ${\begin{cases} x = 3 \land y^{2} = 4 \\ x = 4 \land y^{2} = 3 \end{cases}$ $t h e n x - y^{6} = 3 - 4^{3} o r 4 - 3^{3}$
	Answered by Dwaipayan Shikari last updated on 19/Jul/20

	$2 (a + \frac{1}{a}) = 8$ ${(a + \frac{1}{a})}^{3} - 3 (a + \frac{1}{a}) = a^{3} + \frac{1}{a^{3}}$ $64 - 12 = \frac{a^{6} + 1}{a^{3}} \Rightarrow 52$
	Answered by OlafThorendsen last updated on 19/Jul/20
	$1) x = 7−y^2 (7−y^2 )^3 +y^6 = 91 7^3 −3.7^2 y^2 +3.7y^4 −y^6 +y^6 = 91 3y^4 −21y^2 +49 = 13 3y^4 −21y^2 +36 = 0 Δ = (−21)^2 −4×3×36 = 9 y^2 = ((−(−21)±3)/(2×3)) = 3 or 4 a) y^2 = 3 x = 7−3 = 4 and y^6 = 27 x−y^6 = 4−27 = −23 b) y^2 = 4 x = 7−4 = 3 and y^6 = 64 x−y^6 = 3−64 = −61 S = {−61;−23}$
	$1) x = 7 - y^{2}$ ${(7 - y^{2})}^{3} + y^{6} = 91$ $7^{3} - 3 . 7^{2} y^{2} + 3.7 y^{4} - y^{6} + y^{6} = 91$ $3 y^{4} - 21 y^{2} + 49 = 13$ $3 y^{4} - 21 y^{2} + 36 = 0$ $Δ = {(- 21)}^{2} - 4 \times 3 \times 36 = 9$ $y^{2} = \frac{- (- 21) \pm 3}{2 \times 3} = 3 or 4$ $a) y^{2} = 3$ $x = 7 - 3 = 4 and y^{6} = 27$ $x - y^{6} = 4 - 27 = - 23$ $b) y^{2} = 4$ $x = 7 - 4 = 3 and y^{6} = 64$ $x - y^{6} = 3 - 64 = - 61$ $S = {- 61; - 23}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum