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Last updated at Jan. 23, 2020 by Teachoo
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Misc 9 Evaluate |โ 8(๐ฅ&๐ฆ&๐ฅ+๐ฆ@๐ฆ&๐ฅ+๐ฆ&๐ฅ@๐ฅ+๐ฆ&๐ฅ&๐ฆ)| Let โ = |โ 8(๐ฅ&๐ฆ&๐ฅ+๐ฆ@๐ฆ&๐ฅ+๐ฆ&๐ฅ@๐ฅ+๐ฆ&๐ฅ&๐ฆ)| Applying R1โ R1 + R2 + R3 = |โ 8(๐ฅ+๐ฆ+๐ฅ+๐ฆ&๐ฆ+๐ฅ+๐ฆ+๐ฅ&๐ฅ+๐ฆ+๐ฅ+๐ฆ@๐ฆ&๐ฅ+๐ฆ&๐ฅ@๐ฅ+๐ฆ&๐ฅ&๐ฆ)| = |โ 8(2x+2y&2x+2y&2x+2y@y&x+y&x@x+y&x&y)| = |โ 8(๐(๐ฑ+๐ฒ)&๐(๐ฑ+๐ฒ)&๐(๐ฑ+๐ฒ)@y&x+y&x@x+y&x&y)| Taking common 2(x + y), from R1 = ๐(๐ฑ+๐ฒ) |โ 8(1&1&1@y&x+y&x@x+y&x&y)| Applying C2โ C2 โ C1 = 2(x+y) |โ 8(1&๐โ๐&1@y&x+yโ๐ฆ&x@x+y&xโxโy&y)| = 2(x+y) |โ 8(1&๐&1@y&x&x@x+y&โy&y)| Applying C3 โC3 โ C1 = 2(x+y) |โ 8(1&0&๐โ๐@y&x&xโy@x+y&โy&yโ(x+y))| = 2(x+y) |โ 8(1&0&๐@y&x&xโy@x+y&โy&โx)| Expanding determinant along R1 = 2(x+y) (1|โ 8(๐ฅ&๐ฅโ๐ฆ@โ๐ฆ&โ๐ฅ)|โ0|โ 8(๐ฆ&๐ฅโ๐ฆ@๐ฅ+๐ฆ&โ๐ฅ)|+0|โ 8(๐ฆ&๐ฅ@๐ฅ+๐ฆ&โ๐ฆ)|) = 2(x+y) (1|โ 8(๐ฅ&๐ฅโ๐ฆ@โ๐ฆ&โ๐ฅ)|โ0+0) = 2(x+y) (1( โ x2 โ ( โy) (x โ y)) ) = 2(x+y) ( โ x2 + y (x โ y)) = 2(x+y) ( โ x2 + xy โ y2) = โ 2(x+y) ( x2 + y2 โ xy) = โ 2(x3+y3) Hence , โ = โ 2(๐ฑ๐+๐ฒ๐) (Using a3 + b3 = (a + b) (a2 + b2 โ ab))
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